Calculate the average atomic mass for X

Answer:be careful and relax

Explanation:

Answer:

Hahaha be careful and relax

the answer to the question is B.UV-B

Answer:

Melting-point apparatus

Answer:

1.17 L of H₂

Explanation:

We'll begin by calculating the number of mole in 2.3 g of Mg. This can be obtained as follow:

Mass of Mg = 2.3 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /molar mass

Mole of Mg = 2.3 / 24

Mole of Mg = 0.096 mole

Next, we shall determine the number of mole of H₂ produced by the reaction of 2.3 g (i.e 0.096 mole) of Mg. This can be obtained as follow:

Mg + 2HCl —> MgCl₂ + H₂

From the balanced equation above,

1 mole of Mg reacted to 1 mole of H₂.

Therefore, 0.096 mole of Mg will also react to produce 0.096 mole of H₂.

Finally, we shall determine volume of H₂ produced from the reaction. This can be obtained as follow:

Number of mole (n) of H₂ = 0.096 mole

Pressure (P) = 2 atm

Temperature (T) = 298 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of H₂ =?

PV = nRT

2 × V = 0.096 × 0.0821 × 298

Divide both side by 2

V = (0.096 × 0.0821 × 298) /2

V = 1.17 L

Therefore, 1.17 L of H₂ were obtained from the reaction.

Answer:

The more electronegative atom in a covalent bond

Answer:

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

Explanation:

Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂

Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)

The complete ionic equation includes all the ions and insoluble species.

Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)

The net ionic equation includes only the ions that participate in the reaction and insoluble species.

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

Answer:

It is equal to the total mass of the products.

Explanation:

Hope this helps :)

Answer:

I used a,b c, d in the equation as substituted coefficients to find the unknown for each element of P, Na, O, H, and I got

P4(s) + 4NaOH(aq) + 2H20(l)---->2Ph3 +2Na2HPO3(aq).

which I think should be the answer.

Answer:

Atmosphere to mmHg Conversion Example. Task: Convert 8 atmospheres to mmHg (show work) Formula: atm x 760 = mmHg Calculations: 8 atm x 760 = 6,080 mmHg Result: 8 atm is equal to 6,080 mmHg.

Explanation:

Answer:

The maximum mass of water that could be produced by the chemical reaction=0.441g

Explanation:

We are given that

Given mass of HBr=5.7 g

Given mass of sodium hydroxide=0.980 g

Molar mass of HBr=80.9 g/ Mole

Molar mass of NaOH=40 g/mole

Molar mass of H2O=18 g/mole

Reaction

[tex]HBr+NaOH\rightarrow H_2O+NaBr[/tex]

Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]

Using the formula

Number of moles of HBr=[tex]\frac{5.7}{80.9}=0.0705 moles[/tex]

Number of moles of NaOH=[tex]\frac{0.980}{40}=0.0245moles[/tex]

Hydrogen bromide is in a great excess and the amount of water produced.

Therefore,

Number of moles of water, n(H2O)=Number of moles of NaOH=0.0245moles

Now,

Mass of water=[tex]n(H_2O)\times Molar\;mass\;of\;water[/tex]

Mass of water=[tex]0.0245moles\times 18=0.441g[/tex]

Hence, the maximum mass of water that could be produced by the chemical reaction=0.441g

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

Zn-64 = 48.63%

Zn-66 = 27.90%

Zn-67 = 4.10%

Zn-68 = 18.75%

Zn-70 = .62%

We need to find the  average atomic mass of the given element. It can be solved as follows :

[tex]A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u[/tex]

So, the average atomic mass of zinc is 65.46 u.

Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

Explanation:

The relation between energy and wavelength is as follows.

[tex]E = \frac{hc}{\lambda}\\[/tex]

This means that energy is inversely proportional to wavelength.

As it is given that energy of a hydrogen atom and a helium atom is same.

Let us assume that [tex]E_{hydrogen} = E_{helium} = E'[/tex]. Hence, relation between their wavelengths will be calculated as follows.

[tex]E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}[/tex]    ... (1)

[tex]E_{helium} = \frac{hc}{\lambda_{helium}}[/tex]         ... (2)

Equating the equations (1) and (2) as follows.

[tex]E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'[/tex]

Thus, we can conclude that if a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

Explanation:

4 tctcgcgcgctctchvvyctctc

Answer: The fractional abundance of lighter isotope is 0.518

Explanation:

Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex]           ......(1)

Let the fractional abundance of Ag-107 isotope be 'x'

Atomic mass = 106.90509 amu

Fractional abundance = x

Atomic mass = 108.9047 amu

Fractional abundance = (1 - x)

Average atomic mass of silver = 107.8682 amu

Plugging values in equation 1:

[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]

Fractional abundance of Ag-107 isotope (lighter) = x = 0.518

Hence, the fractional abundance of lighter isotope is 0.518

Answer:

O B. Alkaline earth metals

Explanation:

Noble gases → 8th column.

Alkali metal → first column.

Halogen → 7th

Answer:

B. Alkaline earth metals

Explanation:

Alkaline-earth metals: The alkaline-earth metals make up Group 2 of the periodic table, from beryllium (Be) through radium (Ra). Each of these elements has two electrons in its outermost energy level, which makes the alkaline earths reactive enough that they're rarely found alone in nature. But they're not as reactive as the alkali metals. Their chemical reactions typically occur more slowly and produce less heat compared to the alkali metals.

Answer:

Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:

element & mass %

phosphorus & 39.18%

sulfur & 60.82%

Write the molecular formula of X.

Explanation:

The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.

Empirical formula calculation:                      

element:              phosphorus                       sulfur

co9mposition:      39.185%                            60.82%

divide with

atomic mass:          39.185/31.0 g/mol           60.82/32.0g/mol

                              =1.26mol                           1.90mol

smallest mole ratio:   1.26mol/1.26mol =1      1.90mol/1.26 mol =1.50

multiply with 2:          2                                         3

Hence, the empirical formula is:

P2S3.

Mass of empirical formula is:

158.0g/mol

Given, molecule has molar mass --- 316.25 g/mol

Hence, the ratio is:

316.25g/mol/158.0 =2

Hence, the molecular formula of the compound is :

2 x (P2S3)

=[tex]P_4S_6[/tex]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Answer:

[tex] This\:may\: help[/tex]

Answer:

(1) Write down the chemical reaction in the form of word equation,keeping reactants on left hand side and products on right hand side.

(2) Write symbol and formula of all reactants and products in word equation. (3) Balance the equation by multiplying the symbols and formula by smallest possible figures.

Answer:

60 q

Explanation:

The conversion factor is 100; so 1 quintal = 100 kilograms. In other words, the value in q multiply by 100 to get a value in kg.

Answer:

Consider the following equilibrium:

2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.

What is Kp?

Explanation:

Given,

[tex]Kc=1.08 * 10^7[/tex]

The relation between Kp and Kc is:

[tex]Kp=Kc * (RT)^d^e^l^t^a^(^n^)[/tex]

Where delta n represents the change in the number of moles.

For the given equation,

The Delta n = Number of moles of products - number of moles of reactants

(2-(2+1))

=-1.

Hence,

Kp=Kc/RT.

Thus,

[tex]Kp=1.08 * 10^7 / 8.314 J.K6-1.mol^-^1 x 973 K\\Kp=1335.06[/tex]

The answer is Kp=1335.06

The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].

Explanation:

The relation between [tex]K_p \& K_c[/tex] is given by:

[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]

Where:

[tex]K_c[/tex] = The equilibrium constant of reaction in terms of concentration

[tex]K_p[/tex] = The equilibrium constant of reaction in terms of partial pressure

R= The universal gas constant

T = The temperature of the equilibrium

[tex]n_g[/tex]= Change in gaseus moles

Given:

An equilibrium reaction, 700°C:

[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g),K_c=1.08\times 10^7[/tex]

To find:

The equilibrium constant in terms of partial pressure, [tex]K_p[/tex].

Solution:

The equilibrium constant of reaction in terms of concentration= [tex]K_c[/tex]

[tex]K_c=1.08\times 10^7[/tex]

The equilibrium constant of reaction in terms of partial pressure =[tex]K_p=?[/tex]

The gaseous moles of reactant side = [tex]n_r= 3[/tex]

The gaseous moles of product side = [tex]n_p= 2[/tex]

The temperature at which equilibrium is given = T

[tex]T = 700^oC+273.15 K=973.15K[/tex]

The change in gaseous mole  = [tex]n_g=n_p-n_r=2-3 = -1[/tex]

[tex]K_p=1.08\times 10^7\times (0.0821 atm L/mol K\times 973.15 K)^{-1}\\K_p=1.35\times 10^5[/tex]

The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].

Learn more equilibrium constants here:

brainly.com/question/9173805?referrer=searchResults

brainly.com/question/5877801?referrer=searchResults

Answer:

Explanation:

hope it heloed

Answer:

2.33 mol C

Explanation:

Step 1: Write the balanced generic chemical equation

3 A ⟶ C + 4 D

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of A to C is 3:1.

Step 3: Calculate the number of moles of C produced from 7 moles of A

We will use the previously established molar ratio.

7 mol A × 1 mol C/3 mol A = 2.33 mol C

Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

Explanation:

Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]

Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g

Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.

[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]

Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

Explanation:

Given:

Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]

To find:

The number of atoms of tantalum in 37.1 milligrams.

Solution:

Mass of tantalum = 37.1 mg

[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]

The number of atoms in 0.0371 grams of tantalum = N

Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]

Then a mass of N atoms of tantalum will be:

[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

Learn more about the unitary method here:

brainly.com/question/24566352

brainly.com/question/17743460

Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

Explanation:

Given: Mass = 5.0 g

Volume = 1 L

Molar mass of sucrose = 342.3 g/mol

Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.

[tex]Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol[/tex]

Hence, concentration of sucrose is calculated as follows.

[tex]Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M[/tex]

Formula used to calculate osmotic pressure is as follows.

[tex]\pi = CRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure

C = concentration

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr[/tex]

Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

Answer:

348 inches³

Explanation:

From our previous knowledge of  units conversion:

We know that 1000 cm³ makes 1 Liter.

Thus, for  a 5.70 L automobile engine in cubic meters will be:

= 5.70 × 1000 cm³

= 5700 cm³

Now, the displacement of the automobile in cubic inches provided that 1 inch = 2.534 cm is:

⇒ 5700× (1/ (2.54)³) in³

= 5700×0.0610 in³

= 347.7 in³

≅ 348 inches³

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

Answer:

a.draw 2,3 dicholoro octane

Explanation:

mag isip ka kung paano hehe

Answer:

In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.

Explanation:

In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.

Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.

This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.