Help pls...THANK YOU!!!
Answer:
cell has vast number of constituents in it various scientist proved different theories and discovered different little constituents like mitochondria, DNA etc
How many stars are in the constellation Andromeda?
hellllp i give brainlyst!
Answer:
13 if Im right?
Explanation:
Calculate the root mean square velocity of gaseous argon atoms at 27 ∘C.
Express the velocity to three significant figures and include the appropriate units.
The root mean square velocity of gaseous argon atoms at 27°C is approximately 0.394 m/s.
To calculate the root mean square (rms) velocity of gaseous argon atoms at 27°C, we can use the equation for rms velocity, which is given by vrms = √(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas. To calculate the rms velocity of gaseous argon atoms at 27°C, we need to convert the temperature to Kelvin. The given temperature of 27°C can be converted to Kelvin by adding 273.15 to it, giving us 300.15 K.
The molar mass of argon (Ar) is approximately 39.95 g/mol. Now, we can substitute the values into the equation for rms velocity:
vrms = √(3RT/M) = √[(3 * 8.314 J/mol·K * 300.15 K) / 39.95 g/mol]
Performing the calculations, we get:
vrms ≈ √(6.2115 J·K/mol / 39.95 g/mol) ≈ √(0.1556 J/g) ≈ 0.394 m/s
The rms velocity represents the average velocity of gas particles and is calculated based on the kinetic theory of gases. It is a useful parameter for understanding the behavior and properties of gases, such as their diffusion rates and energy distributions.
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In the electrolysis of molten mixture of NaCl and CaBr2, what is the product that forms at the anode and the cathode?
At the anode, chlorine gas (Cl2) is formed, while at the cathode, sodium metal (Na) is formed in the electrolysis of a molten mixture of NaCl and CaBr2.
In the electrolysis of a molten mixture of NaCl and CaBr2, the products formed at the anode and cathode depend on the individual reactions occurring at each electrode. At the anode, chloride ions (Cl-) from NaCl and bromide ions (Br-) from CaBr2 are discharged. Chloride ions are more easily oxidized than bromide ions, so they tend to be preferentially discharged. At the anode, chloride ions lose electrons and form chlorine gas (Cl2). This reaction is represented as: 2Cl- (aq) → Cl2 (g) + 2e-
At the cathode, sodium ions (Na+) from NaCl and calcium ions (Ca2+) from CaBr2 are discharged. Sodium ions are more easily reduced than calcium ions, so they are preferentially discharged. At the cathode, sodium ions gain electrons and form sodium metal (Na). This reaction is represented as: 2Na+ (aq) + 2e- → 2Na (l)
Calcium ions (Ca2+) from CaBr2 are not reduced at the cathode due to their higher reduction potential compared to sodium ions.
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On Earth a package weighs 19.6 newtons. What is the mass of this package on Earth?
g = 19.6 N/2.2 kg. g = 8.9 m/s2. 7.
The weight an object experience is the product of its mass and gravity. Hence, the package which weighs 19.6 Newtons in earth has a mass of 2 Kg.
What is gravitational force?Gravitational force is a kind of force by which an object attracts other objects into its center of mass. We are all standing on earths surface because of the gravitational pull by earth.
The gravitational force acting on an object depends on its mass and the distance to the object. Therefore, the gravity experienced by each object with different masses or placed at different distances will be different at all.
The weight that's an object have in earth is the product of its mass which is a constant and the acceleration due to gravity that is 9.8 m/s². Thus the object weighing 19.6 has the mass:
mass = 19.6 /9.8
= 2 Kg.
Where one newton is one Kg m/s². Hence, the mass of the package in earth is 2 Kg.
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Which of the following pairs of elements is/are likely to form an ionic bond? Choose one or more: A. Cl and Br B. Na and F C. ca and Cl D. Hand H E. N and C F. Kand I
The pairs of elements that are likely to form an ionic bond are: B. Na and F (Sodium and Fluorine). C. Ca and Cl (Calcium and Chlorine). F. K and I (Potassium and Iodine)
Ionic bonds occur between elements with a large difference in electronegativity. Electronegativity is the tendency of an atom to attract electrons towards itself in a chemical bond.
In options B, C, and F, the pairs of elements have a significant electronegativity difference, which suggests the likelihood of ionic bonding.
In option B, Sodium (Na) has a low electronegativity, while Fluorine (F) has a high electronegativity. The large difference in electronegativity allows Fluorine to attract the electron from Sodium, forming an ionic bond.
In option C, Calcium (Ca) has a low electronegativity, and Chlorine (Cl) has a higher electronegativity. Again, the significant electronegativity difference leads to the formation of an ionic bond.
In option F, Potassium (K) has a low electronegativity, and Iodine (I) has a higher electronegativity, indicating the potential for an ionic bond.
Ionic bonds are characterized by the transfer of electrons from one element to another, resulting in the formation of charged ions that are attracted to each other.
Therefore, the pairs of elements likely to form an ionic bond are B. Na and F, C. Ca and Cl, and F. K and I.
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the molecular weight of average polypropylene chains with a degree of polymerization of 6500, is: group of answer choices 42 amu/mer 182,000 g/mole 273,000 g/mole 273 amu/chain
the molecular weight of average polypropylene chains with a degree of polymerization of 6500 is 273,000 g/mol.
To calculate the molecular weight of polypropylene chains, we need to know the molecular weight of a single monomer unit (mer) and the degree of polymerization.
Given:
Degree of polymerization = 6500
The molecular weight of a single monomer unit of polypropylene (mer) is approximately 42 g/mol. Therefore, we can calculate the molecular weight of the polymer chains as follows:
Molecular weight of polypropylene chains = Molecular weight of a single monomer unit × Degree of polymerization
Molecular weight of polypropylene chains = 42 g/mol × 6500
Molecular weight of polypropylene chains = 273,000 g/mol
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determine the number of moles of oxygen in 454.3 g of sodiuum hydrogen phosphate
15.12 is the number of moles of oxygen in 454.3 g of sodium hydrogen phosphate ([tex]NaH{2}PO_{4}[/tex])
To determine the number of moles of oxygen in 454.3 g of sodium hydrogen phosphate ([tex]NaH{2}PO_{4}[/tex]), we need to use the molar mass and the formula of the compound.
The molar mass of [tex]NaH{2}PO_{4}[/tex] can be calculated by adding up the atomic masses of each element in the formula:
Na: 22.99 g/mol
H: 1.01 g/mol (2 hydrogen atoms)
P: 30.97 g/mol
O: 16.00 g/mol (4 oxygen atoms)
Molar mass of [tex]NaH{2}PO_{4}[/tex] = (22.99 g/mol) + (1.01 g/mol × 2) + 30.97 g/mol + (16.00 g/mol × 4) = 120.00 g/mol
Now, we can use the molar mass to calculate the number of moles of [tex]NaH{2}PO_{4}[/tex]:
Number of moles = Mass of substance / Molar mass
Number of moles of [tex]NaH{2}PO_{4}[/tex] = 454.3 g / 120.00 g/mol ≈ 3.78 moles
Since there are four oxygen atoms in one molecule of [tex]NaH{2}PO_{4}[/tex], we can calculate the number of moles of oxygen by multiplying the number of moles of [tex]NaH{2}PO_{4}[/tex] by the ratio of oxygen atoms to [tex]NaH{2}PO_{4}[/tex] molecules:
Number of moles of oxygen = 3.78 moles × 4 = 15.12 moles
Therefore, there are approximately 15.12 moles of oxygen in 454.3 g of sodium hydrogen phosphate.
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At STP, a 50.-gram sample of H2O(l) and a 100.-gram sample of H2O(1) have
1. the same chemical properties
2. the same volume
3.different temperatures
4.different empirical formulas
Answer:
The same chemical properties
Explanation:
Different masses of water at the same temp and pressure will have different volumes and the same empirical formulas and the same chemical properties
At STP, a 50.-gram sample of H2O(l) and a 100.-gram sample of H2O(l) have the same chemical properties.
The temperature, volume and other physical properties of a sample may vary with the mass of substance present.
However, the chemical properties of a substance do not change irrespective of the amount of substance present. Even the smallest mass of water still retains all the chemical properties of water.
Therefore, at STP, a 50.-gram sample of H2O(l) and a 100.-gram sample of H2O(l) have the same chemical properties.
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If there were only three electron groups around an atom, how would they be
arranged?
A. They would be arranged in a trigonal pyramidal orientation.
B. They would be arranged in a trigonal planar orientation.
C. They would be arranged in a bent orientation.
D. They would be arranged in a linear orientation.
Explanation:
C they would be arranged in a bent Orientation
Answer:
B
Explanation:
They would be arranged in a trigonal planar orientation.
Daniel is trying to determine if one of Newton’s ideas is a law or a theory. The idea that for every action there is an equal but opposite reaction is a fact, based upon numerous experiments, that has never been revised or changed. The idea is based on observations but is not an explanation of those observations. Daniel should determine that this is a law, because it is a fact that does not change. this is a law, because is based on observations. this is a theory, because it does not explain observations. this is a theory, because it is based on numerous experiments.
true
Explanation:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... The third law states that for every action (force) in nature there is an equal and opposite reaction.
Answer:
This is a law because it is based on observations (Newton's Third Law)
Explanation:
5. A beaker is filled with water and has an initial volume of 100 ml. After an object is placed into
the beaker, the volume of the beak measures 150 ml. If the mass of the object is 200 grams,
what is the density of the object?
Answer:
4
Explanation:
Mass divided by Volume = Density
The objects mass is 200 grams and the volume of the water displaced is 50
So 200 divided by 50 equals 4
.Which one of the following elements exhibit maximum number of valence electrons?
(a) Na
(b) Al
(c) Si
(d) P
The element that exhibits the maximum number of valence electrons among the given options is (c) Si (silicon).
A) Valence electrons are the electrons in the outermost energy level (valence shell) of an atom, which determine its chemical properties and reactivity. The maximum number of valence electrons an element can have is determined by the electron configuration and the number of electrons in its outermost energy level.
Let's examine the electron configurations of the given elements:
(a) Sodium (Na) - 1s² 2s² 2p⁶ 3s¹
(b) Aluminum (Al) - 1s² 2s² 2p⁶ 3s² 3p¹
(c) Silicon (Si) - 1s² 2s² 2p⁶ 3s² 3p²
(d) Phosphorus (P) - 1s² 2s² 2p⁶ 3s² 3p³
From the electron configurations, we can determine the number of valence electrons by considering the electrons in the outermost energy level (highest principal quantum number).
In the case of silicon (Si), the electron configuration shows that it has four electrons in its outermost energy level (3s² 3p²). Therefore, silicon exhibits the maximum number of valence electrons among the given options.
Hence, the element that exhibits the maximum number of valence electrons is (c) Si (silicon).
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Definition of Magnetism
Answer:
Magnetism is the force of attraction on a magnetic substance by a magnet.
Explanation:
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Calculate the pH for each of the cases in the titration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 M HBr(aq). The Kb of pyridine is 1.7×10−9.
before addition of any HBr pH=
after addition of 12.5 mL of HBr pH=
after addition of 23.0 mL of HBr pH=
after addition of 25.0 mL of HBr pH=
after addition of 34.0 mL of HBr pH=
Before any HBr is added, the pH is approximately 5.23. After the addition of HBr, the pH becomes 1.00, which remains constant throughout subsequent additions of HBr.
To calculate the pH at different stages of the titration between pyridine (C5H5N) and HBr, we need to consider the reaction between the weak base pyridine and the strong acid HBr. Pyridine acts as a base and accepts a proton (H+) from HBr, forming the conjugate acid of pyridine (C5H5NH+).
Before any HBr is added:
Since pyridine is a weak base, we can consider the initial solution as a basic solution. The pH can be calculated using the pKb of pyridine:
pKb = -log(Kb) = -log(1.7×10^(-9)) ≈ 8.77
pH = 14 - pKb ≈ 14 - 8.77 ≈ 5.23
After addition of 12.5 mL of HBr:
The volume of the solution has increased to 25.0 mL + 12.5 mL = 37.5 mL. We can assume that the HBr completely reacts with pyridine, forming its conjugate acid. Since HBr is a strong acid, we can consider the solution as acidic.
pH = -log[H+] = -log(0.100 M) = 1.00
After addition of 23.0 mL of HBr:
The volume of the solution is now 25.0 mL + 23.0 mL = 48.0 mL. Again, we assume complete reaction of HBr with pyridine.
pH = -log[H+] = -log(0.100 M) = 1.00
After addition of 25.0 mL of HBr:
The volume of the solution is now 25.0 mL + 25.0 mL = 50.0 mL. Complete reaction of HBr with pyridine is assumed.
pH = -log[H+] = -log(0.100 M) = 1.00
After addition of 34.0 mL of HBr:
The volume of the solution is now 25.0 mL + 34.0 mL = 59.0 mL. Complete reaction of HBr with pyridine is assumed.
pH = -log[H+] = -log(0.100 M) = 1.00
In summary:
- Before any HBr is added: pH ≈ 5.23
- After addition of 12.5 mL of HBr: pH = 1.00
- After addition of 23.0 mL of HBr: pH = 1.00
- After addition of 25.0 mL of HBr: pH = 1.00
- After addition of 34.0 mL of HBr: pH = 1.00
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Physical or Chemical Changes
Gallium melting into its liquid form (at 30°C) is an example of a
Type text here.
change, where some
properties of a material change, but the composition of the material
does not change. Physical changes can be described as reversible or
irreversible, with matter state changes being an example of a(n)
Type text here.
physical change.
Answer: 1. physical 2. Reversible
Gallium melting into its liquid form (at 30°C) is an example of a physical change and they're reversible.
Gallium melting into its liquid form (at 30°C) is an example of a physical change, where some properties of a material change, but the composition of the material does not change. Physical changes can be described as reversible or irreversible, with matter state changes being an example of a reversible physical change.It should be noted that the appearance of a substance only changes in the physical change but the chemical composition doesn't change.In conclusion, chemical changes bring about a change in substance into a substance that's entirely different.
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write a balanced reaction equation for the dehydration of cyclohexanol
Dehydration is the process of eliminating water from the given compound. When cyclohexanol dehydrates, it forms cyclohexene as written here:
C₆H₁₁OH ⇒[tex]C_6H_{10} + H_2O\\[/tex]
What is cyclohexanol?
Cyclohexanol is an alcohol formed from the cyclic alkane cyclohexane and water or other hydroxyl groups. The chemical formula of this compound is
C₆H₁₁OH.
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The osmotic pressure of a solution containing 22.7 mg of an unknown protein in 50.0 mL of solution is 2.88 mmHg at 25 degreeC. Determine the molar mass of the protein. a) 246 g/mol b) 3.85 g/mol c) 2.93 times 10^3 g/mol d) 147 g/mol
The molar mass of the protein is approximately 74,426 g/mol. Since none of the provided answer choices match exactly, it seems there might be a mistake in the options given.
To determine the molar mass of the protein, we can use the formula for osmotic pressure:
π = (n/V)RT
Where π is the osmotic pressure, n is the number of moles of solute, V is the volume of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.
Given that the osmotic pressure is 2.88 mmHg, we need to convert it to atm by dividing it by 760 mmHg/atm:
π = 2.88 mmHg / 760 mmHg/atm = 0.00379 atm
The volume of the solution is 50.0 mL, which needs to be converted to liters:
V = 50.0 mL / 1000 mL/L = 0.0500 L
The temperature is given as 25 degrees Celsius, which needs to be converted to Kelvin:
T = 25 degrees Celsius + 273.15 = 298.15 K
Plugging these values into the osmotic pressure equation and rearranging to solve for the number of moles of solute:
n = (πV) / (RT)
n = (0.00379 atm * 0.0500 L) / (0.0821 L·atm/(mol·K) * 298.15 K)
n = 0.000305 mol
The molar mass of the protein can be calculated by dividing the mass of the protein by the number of moles:
Molar mass = mass / moles
Molar mass = 22.7 mg / 0.000305 mol
Molar mass = 74,426 g/mol
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How many grams of 50 wt% NaOH (FM 40.00) should be diluted to 1.00 L to make 0.10 M NaOH? (Answer with two digits.)
4.00 grams of 50 wt% NaOH should be diluted to 1.00 L to make a 0.10 M NaOH solution.
To calculate the amount of 50 wt% NaOH needed to prepare a 0.10 M NaOH solution, we need to know the molar mass of NaOH and the desired volume of the solution. The molar mass of NaOH is approximately 40.00 g/mol.
The given solution is 50 wt% NaOH, which means that 50 grams of NaOH are present in every 100 grams of solution. We need to find out how many grams of NaOH are needed to make 1.00 L of 0.10 M NaOH solution. To calculate the amount of NaOH needed, we can use the formula:
Amount (in moles) = concentration (in M) × volume (in liters)
The desired concentration is 0.10 M and the volume is 1.00 L. Substituting the values into the formula:
Amount (in moles) = 0.10 mol/L × 1.00 L = 0.10 mol
To convert moles to grams, we multiply the amount (in moles) by the molar mass:
Mass = amount (in moles) × molar mass
Mass = 0.10 mol × 40.00 g/mol = 4.00 g
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How many molecules of CO2 are contained in a 5.00 L tank at 7.53 atm and 485 K? a. 9.45 x 10^23 molecules b. 4.16 x 10^24 molecules c. 5.69 x 10^23 molecules d. None of the above e. 2.45 x 10^24 molecules
The number of molecules of CO2 contained in a 5.00 L tank at 7.53 atm and 485 K can be calculated using the ideal gas law and Avogadro's number. The correct answer is option e, which is 2.45 x 10^24 molecules.
To determine the number of CO2 molecules in the tank, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure in atmospheres (7.53 atm)
V = Volume in liters (5.00 L)
n = Number of moles of gas (to be determined)
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature in Kelvin (485 K)
Rearranging the equation to solve for n:
n = PV / RT
Substituting the given values:
n = (7.53 atm) * (5.00 L) / (0.0821 L·atm/(mol·K) * 485 K)
n = 0.1848 mol
Now, to convert moles to molecules, we can use Avogadro's number, which states that 1 mole of a substance contains 6.022 x 10^23 molecules.
Number of molecules = (0.1848 mol) * (6.022 x 10^23 molecules/mol)
Number of molecules = 1.112 x 10^23 molecules
Therefore, the correct answer is option e, which is 2.45 x 10^24 molecules.
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Are the eight valence electrons the most stable configuration?
Answer: No, the noble gases are
Explanation:
Answer:
The most stable electron configuration is that of a noble gas, due to the fact that its valence shell is filled. For helium, that means two valence electrons (a duet) in the 1s sublevel, and for the rest it means eight valence electrons (an octet) in the outermost s and p sublevels.
bromine is a liquid at a room temperature. the volume of the sample of bromine is measured in a 50 mL beaker and a 100 mL beaker. how will the two measurements compare?
Answer:
This question is incomplete
Explanation:
This question is incomplete but generally it seem to test the knowledge of accuracy. Accuracy is the closeness of a measured value to the specific or actual value of the substance being measured. Hence, the smaller the entire scale that can contain all the substance being measured, the higher the possibility of accuracy to be achieved. For example, a pen of 10.4 cm in length can be better/accurately measured (length) with a 30 cm meter rule than a 1 m meter rule.
From the question, the 50 mL beaker will provide a "more accurate" measurement of the volume of the bromine than the 100 mL beaker if the 50 mL beaker can contain all the sample of bromine water provided.
to what volume should you dilute 0.400 l of a 13.0 mnaoh solution to obtain a 3.00 mnaoh solution? express your answer with the appropriate units.
We can use the M1V1 = M2V2 equation to calculate the volume required to dilute 0.400 L of a 13.0 M NaOH solution to obtain a 3.00 M NaOH solution.M1V1 = M2V2 equation,M1V1 = M2V2V1 = (M2 × V2)/M1
In this scenario:M1 = 13.0 M, V1 = 0.400 L, M2 = 3.00 M. We can put these values into the M1V1 = M2V2 equation and solve for
V2:V2 = (M1 × V1)/M2V2 = (13.0 M × 0.400 L)/3.00 MV2 = 1.73 L
To determine the volume required to dilute 0.400 L of a 13.0 M NaOH solution to a 3.00 M NaOH solution, we can use the M1V1 = M2V2 equation. The formula tells us that the initial concentration and volume, as well as the final concentration and volume, are proportional.If we look at the equation, we can observe that the final volume (V2) is proportional to the initial volume (V1) and the ratio of the initial and final concentrations (M1/M2). We substitute the values given in the problem into the equation and solve for V2:V2 = (M1 × V1)/M2V2 = (13.0 M × 0.400 L)/3.00 MV2 = 1.73 L Therefore, to dilute 0.400 L of a 13.0 M NaOH solution to a 3.00 M NaOH solution, 1.73 L of water must be added to the solution.
We can conclude that to prepare a 3.00 M NaOH solution from a 13.0 M NaOH solution, 1.73 L of water should be added to 0.400 L of 13.0 M NaOH solution.
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What happens to how much space the water fills when it changes into water vapor?
A.
the water vapor fills less space
B.
the water vapor fills more space
C.
the water vapor fills the same space
D.
the water vapor does not fill any space
Answer:
regr vldfpelrg
Explanation:
Answer:
i think A is the answer (the water vapor fills less space)
Please help me answer this it’s due today. I will give brainliest
Answer: X should represent H, hydrogen.
Explanation:
The H is the only one that hasnt been stated in the left side of the formula. H has three atoms as well.
an unknown metal cylinder was placed in a 100 ml graduated cylinder containing 25. 2 ml of water. the water increased to 48.3 ml. the cylinder weighed 101.356 g. calculate the density of the metal.
The first step is to calculate the volume of the metal cylinder. This can be found by subtracting the final volume of water (48.3 mL) from the initial volume of water (25.2 mL), giving a volume of 23.1 mL. This volume is equivalent to 23.1 cm³ because 1 mL = 1 cm³.
The next step is to calculate the density of the metal. We know the mass of the metal cylinder (101.356 g), so we can divide this by its volume (23.1 cm³): density = mass/volume density = 101.356 g/23.1 cm³density = 4.39 g/cm³. We were given the mass of an unknown metal cylinder as 101.356 g and the initial volume of water as 25.2 mL. Upon immersion of the metal cylinder in the graduated cylinder, the water level rose to 48.3 mL. To determine the density of the metal cylinder, we need to find the volume of the cylinder. We can calculate the volume of the metal cylinder by finding the difference between the final and initial volumes of water. This gives us a volume of 23.1 mL. This volume is equivalent to 23.1 cm³ since 1 mL is equivalent to 1 cm³.To calculate the density of the metal cylinder, we divide the mass of the cylinder by its volume. This gives us a density of 4.39 g/cm³. Therefore, the unknown metal cylinder has a density of 4.39 g/cm³.
The density of the metal cylinder can be found by dividing its mass by its volume. In this case, we were given the mass of the cylinder and the initial and final volumes of water, so we used the difference between these volumes to find the volume of the cylinder. Dividing the mass by the volume gave us a density of 4.39 g/cm³ for the metal cylinder.
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Find the distance between these points.
W(-6, -8), X(6, 8)
Distance =
[tex] \sqrt{ {(6 + 6)}^{2} + {(8 + 8)}^{2} } = \sqrt{144 + 256} = \sqrt{400} = 20[/tex]
zinc reacts with hydrochloric acid according to the reaction equation zn(s) 2hcl(aq)⟶zncl2(aq) h2(g) how many milliliters of 3.50 m hcl(aq) are required to react with 4.75 g zn(s)?
Approximately 67.8 mL of 3.50 M HCl(aq) is required to react with 4.75 g of Zn(s).
To calculate the volume of 3.50 M HCl(aq) required to react with 4.75 g of Zn(s), we need to use the stoichiometry of the balanced chemical equation.
By converting the mass of Zn to moles using its molar mass, and then using the mole ratio between Zn and HCl, we can determine the volume of HCl required. In this case, the volume of 3.50 M HCl(aq) needed is approximately 67.8 mL.
First, we need to determine the number of moles of Zn(s) using its molar mass. The molar mass of Zn is 65.38 g/mol:
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 4.75 g / 65.38 g/mol
Next, we use the balanced chemical equation to establish the mole ratio between Zn and HCl. From the equation Zn(s) + 2HCl(aq) ⟶ ZnCl2(aq) + H2(g), we can see that the ratio is 1:2:
moles of HCl = 2 * moles of Zn
Now, we can calculate the volume of 3.50 M HCl(aq) needed using the concentration and the moles of HCl:
moles of HCl = volume of HCl * concentration of HCl
Rearranging the equation to solve for the volume:
volume of HCl = moles of HCl / concentration of HCl
Substituting the known values:
volume of HCl = (2 * moles of Zn) / 3.50 M
Calculating the volume:
volume of HCl = (2 * 4.75 g / 65.38 g/mol) / 3.50 M
volume of HCl ≈ 67.8 mL
Therefore, approximately 67.8 mL of 3.50 M HCl(aq) is required to react with 4.75 g of Zn(s).
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eparate the pennies you have into two groups: one group with pennies dated before 1982 and the other with pennies dated after 1982. Compare the two groups. Record any similarities or differences in size or appearance that you notice between the two groups.
The comparison between the two groups of pennies reveals distinct differences in their composition, appearance, and weight.
Upon separating the pennies into two groups based on their dates, I observed notable differences between the two groups. Pennies dated before 1982 are primarily composed of copper, while those dated after 1982 are made of zinc with a thin copper plating.
In terms of appearance, the pre-1982 pennies have a reddish-brown color due to their high copper content. They often show signs of aging, such as discoloration, tarnish, and wear. In contrast, the post-1982 pennies have a brighter and shinier appearance, resembling a silver-like hue due to the copper coating.
In terms of size, both groups of pennies have the same diameter and thickness. However, the pre-1982 pennies tend to be slightly heavier due to the higher density of copper compared to zinc, which is used in post-1982 pennies.
Pennies dated before 1982 are made of copper, have a reddish-brown color, and are slightly heavier. Pennies dated after 1982, on the other hand, are made of zinc with a copper coating, appear brighter and more silver-like, and are slightly lighter.
These differences arose from the change in materials used by the U.S. Mint in 1982 to reduce production costs.
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how much energy is needed to raise 70 g of paper 40 degrees celsius?
Explanation: Heat capacity is the amount of heat required to change the temperature of a ... Therefore, specific heat is measured in Joules per g times degree Celsius
