the molarity of the final solution is 0.045 M If 60. 0 ml of a 1. 5 m hcl solution is put into a flask and diluted with water to make 2000 ml of solution.
To solve this problem, we can use the formula:
M1V1 = M2V2
Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
We are given that the initial volume is 60.0 ml and the initial molarity is 1.5 M. We are also given that the final volume is 2000 ml (which is the sum of the initial volume and the volume of water added).
Using the formula above, we can solve for the final molarity:
M1V1 = M2V2
(1.5 M) (60.0 ml) = M2 (2000 ml)
M2 = (1.5 M) (60.0 ml) / (2000 ml)
M2 = 0.045 M
Therefore, the molarity of the final solution is 0.045 M.
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The cytoplasm of a cell is filled with salts, sugars, proteins, and other molecules.
This means that cells usually have a lower concentration of water compared with
their external environment.
Considering this, what would happen if a cell were placed in
freshwater?
A. The cell would shrink as water moved along a concentration gradient from
low to high.
B. The cell would remain the same due to its semipermeable membrane.
C. The cell would shrink as water moved from an area of high solute
concentration to an area of low solute concentration.
D. The cell would burst as water moved into the cell along a concentration
gradient from high to low.
The cell would burst as water moved into the cell along a concentration gradient from high to low.
What would happen to a salt filled cell in fresh water?If a salt-filled cell is placed in fresh water, osmosis will occur. Osmosis is the diffusion of water molecules from an area of high concentration to an area of low concentration through a semi-permeable membrane. In this case, the semi-permeable membrane is the cell membrane that surrounds the salt-filled cell.
Since the concentration of salt inside the cell is higher than the concentration of salt in the fresh water, water molecules will move from the fresh water into the cell in an attempt to balance the concentration of salt on both sides of the membrane. This results in an increase in the volume of the cell, which can cause the cell to swell or even burst.
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If the reaction yield is 95.7% how many grams of lead oxide will be produced by the decomposition of 2.50g of lead carbonate
If the reaction yield is 95.7%, then 2.37 grams of lead (II) oxide will be produced by the decomposition of 2.50 grams of lead (II) carbonate.
Answer:
2.54g
Explanation:
Lead carbonate (PbCO3) can be decomposed to produce lead oxide (PbO) and carbon dioxide (CO2). The equation for this reaction is as follows:
PbCO3(s) -> PbO(s) + CO2(g)
Let's assume that the reaction yield is 95.7%, meaning that 95.7% of the theoretical amount of lead oxide produced in the reaction is actually obtained. To find the actual amount of lead oxide produced, we first need to find the theoretical amount of lead oxide produced by the reaction.
The theoretical amount of lead oxide can be calculated using the stoichiometry of the reaction, where the number of moles of reactants is balanced with the number of moles of products. If we assume that 2.50g of lead carbonate is decomposed, the number of moles of lead carbonate can be calculated as follows:
n = m/M
where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance. For lead carbonate, the molar mass is:
M = 207.19 g/mol
So, the number of moles of lead carbonate is:
n = 2.50 g / 207.19 g/mol = 0.01202 mol
Since the reaction is balanced, the number of moles of lead oxide produced should be equal to the number of moles of lead carbonate. The mass of lead oxide produced can be calculated using the number of moles and the molar mass of lead oxide:
m = n x M
where M = 223.20 g/mol is the molar mass of lead oxide. So, the mass of lead oxide produced is:
m = 0.01202 mol x 223.20 g/mol = 2.68 g
Since the reaction yield is 95.7%, the actual amount of lead oxide produced is:
actual_mass = 0.957 x 2.68 g = 2.54 g
So, approximately 2.54g of lead oxide will be produced by the decomposition of 2.50g of lead carbonate.
if u coated the surface of a penny with a thin layer of a nonpolar liquid (oil) and then dropped alcohol and water onto the penny what would occur?
If you coated the surface of a penny with a thin layer of a nonpolar liquid (oil) and then dropped alcohol and water onto the penny what would occur is that the alcohol will mix with the water and spread out to form a thin layer on top of the oil.
What are non-polar substances?Any molecule without any electrical charges or partial charges is said to be a nonpolar substance. Atoms with similar levels of electronegativity make up nonpolar compounds.
Alcohol is a polar substance and water is also a polar substance.
Hence, the water and alcohol will not mix with the oil because oil is nonpolar and will not dissolve in a polar solvent like water or alcohol.
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Which of the following is NOT a assumption of the kinetic theory of gases
Answer:
All molecules have same speed.
Explanation:
Molecules of an ideal gas move randomly with different speeds.
Zinc metal and aqueous copper (II) chloride react to produce aqueous zinc chloride and
copper metal.
O Combustion
O Synthesis
O Single Replacement
O Decomposition
O Double Replacement
O Acid/Base Neutralization
Answer: Single Replacement
Explanation:
This reaction is single replacement because the Zinc pushes out the Copper from the CuCl2 to form ZnCl. This happens because Zinc is above Copper in the activity series.
Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO_2 (aq) + CO(aq) HbCO(aq) + O_2(aq) a. Use the following reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction. Hb(aq) + O_2(aq) HbO_2(aq) K_c = 1.8 Hb(aq) + CO(aq) HbCO(aq) K_c = 306 b. Suppose that an air mixture becomes polluted with carbon monoxide at a level of 0.10%. Assuming the air contains 20.0% oxygen, and that the oxygen and carbon monoxide ratios that dissolve in the blood are identical to the ratios in the air, what is the ratio of HbCO to HbO2 in the bloodstream? Comment on the toxicity of carbon monoxide.
a. We may use the equilibrium constants for the two reactions provided to get the equilibrium constant for the reaction HbO2(aq) + CO(aq) -> HbCO(aq) + O2(aq). O2(aq) + Hb(aq).
= 1.8 Hb(aq) + CO(aq) = HbO2(aq) Kc Kc HbCO(aq) = 306 Considering the forward and reverse. reaction ratios for the whole response, we get: ([HbCO][O2])/ K eq ([HbO2][CO]) Substituting in the equilibrium constants for the two specified reactions yields: K eq = (306 * [O2]) / 1.8 K eq = (306 * [O2]) / 1.8 The concentration of O2 in the circulation at body temperature is around 0.13 mM, or 0.13/1000 moles per liter. Substituting this value yields: (306 * 0.13/1000) / 1.8 = 0.022 K eq At body temperature, the equilibrium constant for the reaction HbO2(aq) + CO(aq) -> HbCO(aq) + O2(aq) is around 0.022. b. If 0.10% of the air mixture includes carbon monoxide and 20.0% oxygen, which indicates that there is one molecule of carbon monoxide and 200 molecules of oxygen in every 1000 molecules of air. Considering that the oxygen and carbon monoxide concentrations in blood are the same as those in air, this means that for every 200 molecules of HbO2 that join with oxygen, 1 molecule of HbO2 will bond with carbon monoxide to generate HbCO. The circulatory HbCO/HbO2 ratio can be computed as follows: [CO]/([O2]/200) = [CO] / ([O2]/200) Substituting the provided values yields: (1/1000) / (20/100 * 200) = [HbCO]/[HbO2 1/20000 [HbCO]/[HbO2] As a result, the circulatory HbCO/HbO2 ratio is 1:20000, or 0.005%. Carbon monoxide is extremely harmful because it bonds to Since it binds to hemoglobin considerably more firmly than oxygen, it can displace oxygen from hemoglobin and prevent it from reaching essential organs. This can result in tissue damage, organ failure, and even death. Low quantities of carbon monoxide can be hazardous, causing headaches, dizziness, nausea, and other symptoms.
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2 NaClO3 → 2 NaCl + 3 O2
Calculate the mass of O2 produced as the result of the decomposition of 843 g of NaClO3.
Taking into account the reaction stoichiometry, 380.12 grams of O₂ are produced as the result of the decomposition of 843 g of NaClO₃.
Reaction stoichiometryIn first place, the balanced reaction iS:
2 NaClO₃ → 2 NaCl + 3 O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
NaClO₃: 2 molesNaCl: 2 molesO₂: 3 molesThe molar mass of the compounds is:
NaClO₃: 106.45 g/moleNaCl: 58.45 g/moleO₂: 32 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
NaClO₃: 2 moles ×106.45 g/mole= 212.9 gramsNaCl: 2 moles ×58.45 g/mole= 116.9 gramsO₂: 3 moles ×32 g/mole= 96 gramsMass of O₂ formedThe following rule of three can be applied: if by reaction stoichiometry 212.9 grams of NaClO₃ form 96 grams of O₂, 843 grams of NaClO₃ form how much mass of O₂?
mass of O₂= (843 grams of NaClO₃× 96 grams of O₂) ÷212.9 grams of NaClO₃
mass of O₂= 380.12 grams
Finally, 380.12 grams of O₂ are formed.
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What are some strategies to improve the powerless language?
Some strategies which are used to improve the powerless language include the reduction of the use of the following below:
hedgesintensifiersformal grammar etc.What is a Powerless language?This is referred to as the type of language which suggests to listeners that you are uncertain, insufficient, and not entirely confident with oneself due ti various reasons.
Examples of the reasons are the use of hedges , intensifiers etc which have negative effects on how the speaker is perceived. It is usually characterized by Hesitation like “Well” or “Um” etc and it should be avoided to improve it thereby making it he correct choice.
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Which crop did texas farmers produce the most of in the years after the civil war?.
In the years after the civil war texas farmers produced cotton the most among all crops,
The Importance of Cotton in Texas Agriculture After the Civil WarIn the years following the Civil War, cotton was the most important cash crop in Texas and the primary source of income for many farmers. Texas had favorable climatic conditions and fertile soils that were well-suited for cotton cultivation. Additionally, the demand for cotton was high both domestically and internationally, which made it a lucrative crop for farmers. As a result, cotton production in Texas increased significantly in the years after the Civil War, with the state becoming one of the largest cotton producers in the country. However, this dependence on a single crop also left farmers vulnerable to market fluctuations and crop failures, which led to economic hardship and debt for many.
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The density of pure iron is 7.874 g/cm³. What is the mass (in grams) of a piece of iron with a volume of 20.00 cm³?
Answer:
157.48 g
Explanation:
To calculate the mass of a piece of iron with a volume of 20.00 cm³, we can use the formula:
mass = density x volume
where density is the density of pure iron (7.874 g/cm³) and volume is the volume of the iron piece (20.00 cm³).
mass = 7.874 g/cm³ x 20.00 cm³ = 157.48 g
So, the mass of the iron piece is 157.48 g.
How many grams of Mn are in 1.07x10^24 atoms
The mass (in grams) of Mn present in 1.07×10²⁴ atoms is 97.735 grams
How do I determine the mass of Mn present?We'll begin by obtaining the number of mole of Mn. This can be obtained as follow:
From Avogadro's hypothesis,
6.02×10²³ atoms = 1 mole of Mn
Thus,
1.07×10²⁴ atoms = (1.07×10²⁴ atoms × 1 mole) / 6.02×10²³ atoms
1.07×10²⁴ atoms = 1.777 moles of Mn
Now, we shall determine the mass (in grams) of Mn. Details below:
Number of mole of Mn = 1.777 moles Molar mass of Mn = 55 g/mol Mass of Mn =?Mole = mass / molar mass
Cross multiply
Mass = Mole × molar mass
Mass of Mn = 1.777 × 55
Mass of Mn = 97.735 grams
Thus, from the above calculation, it is evident that the mass of Mn is 97.735 grams
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