All categories

3 years ago

Given 6193 mL of a gas at 62.3 °C. What is its volume at 38.1 °C?

Chemistry

1 answer:

Luden [163]3 years ago

5 0

Answer:

5746.0 mL.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.

V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.

∴ V₂ = V₁T₂/T₁ = (6193.0 mL)(311.1 K)/(335.3 K) = 5746.0 mL.

You might be interested in

Why are only certain colors absorbed by and emitted from an atom?

zavuch27 [327]

The way I would explain it is quite difficult to understand, so this is what Google says. "The wavelength (or equivalently, frequency) of the photon is determined by the difference in energy between the two states. These emitted photons form the element's spectrum. The fact that only certain colors appear in an element's atomic emission spectrum means that only certain frequencies of light are emitted." I hope this helped.

6 0

4 years ago

In poor quality fireworks what will you notice and why

anastassius [24]

The best answer I could find was when you Google it, that the fuse is of poor quality. I cannot leave you a link, but you can find it for yourself. Put in poor quality fireworks and all sorts of things will pop up. No pun intended.

8 0

3 years ago

Read 2 more answers

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

I need help with problem 103 all of the questions.

AfilCa [17]

A. is decomposition so HCL = H2 + Cl2

not balanced cause hcl needs 2

2HCL = H2 + Cl2

balanced

b. Br2 + Al-i = AlBr3 + I2 single rep.

not balanced since br need 3 so watch carefully cause many changes needed

3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2

3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there

3Br2 + 2Ali = 2AlBr3 + I2

Balanced

C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2

2Na + S = Na2S balanced.

4 0

3 years ago

Which transition metals usually form only one monatomic cation? select all that apply.

lyudmila [28]

Zn, Cd, and Ag are transition metals that usually form only one monoatomic cation.

A monatomic cation is a cation made of only one atom.

Cations are positively charged ions, in this example Ag⁺, Cd²⁺ and Zn²⁺.

These cations form only one type of ion, while iron and copper form more than one type of cations.

Iron and copper form cations with different charges (Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).

It depends on electron configuration which type would be formed.

Electron configuration of zinc atom: ₃₀Zn 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²

Transition metals are elements in the d-block of the Periodic table.

More about transition metals: brainly.com/question/12843347

#SPJ4

5 0

2 years ago