Ask question

Ask question

All categories

3 years ago

6

Calculate the volume of helium gas sample that has a mass of 1.2 g at room conditions

1 answer:

Diano4ka-milaya [45]3 years ago

3 0

1.2over2=0.6 as molarity then we know that molar volume=22.4dm3
molarity=volume per molar volume
therefore volume=molarity ×molar volume= 0.6 x 22.4= 13.44

You might be interested in

The average speeds of gas molecules in cylinders A, B, C, and D are 0.001 m/s, 0.05 m/s, 0.1 m/s, and 0.5 m/s respectively. Whic

MrRissso [65]

Answer:

cylindar a

Explanation:

I took the test

7 0

3 years ago

Read the description of how ancient metalworkers improved their iron. Then identify the solute and solvent by filling in the bla

slega [8]

Answer:iron and carbon

Explanation:I took the test it’s iron and carbon

6 0

4 years ago

Read 2 more answers

In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to

Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

Explanation:

The measured pH of the solution = 5.153

$C_6H_5OH\rightarrow C_6H_5O^-+H^+$

Initially c

At eq'm c-x x x

The expression of dissociation constant is given as:

$K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}$

Concentration of phenoxide ions and hydrogen ions are equal to x.

$pH=-\log[x]$

$5.153=-\log[x]$

$x=7.03\times 10^{-6} M$

$K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}$

$K_a=9.34\times 10^{-11}$

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

4 0

4 years ago

Calculate ΔG∘rxn for the following reaction:4CO(g)+2NO2(g)→4CO2(g)+N2(g).Use the following reactions and given ΔG∘rxn values:A)

mihalych1998 [28]

Answer : The value of $\Delta G^o_{rxn}$ for the reaction is -1131.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The given main reaction is,

$4CO(g)+2NO_2(g)\rightarrow 4CO_2(g)+N_2(g)$ $\Delta G^o_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2NO(g)+O_2(g)\rightarrow 2NO_2(g)$ $\Delta G^o_1=-72.6kJ$

(2) $2CO(g)+O_2(g)\rightarrow 2CO_2(g)$ $\Delta G^o_2=-514.4kJ$

(3) $\frac{1}{2}O_2(g)+\frac{1}{2}N_2(g)\rightarrow NO(g)$ $\Delta G^o_3=87.6kJ$

Now we will reverse the reaction 1, multiply the reaction 2 by 2, reverse and half the reaction 3 and then adding all the equations, we get :

(1) $2NO_2(g)\rightarrow 2NO(g)+O_2(g)$ $\Delta G^o_1=72.6kJ$

(2) $4CO(g)+2O_2(g)\rightarrow 4CO_2(g)$ $\Delta G^o_2=2\times (-514.4kJ)=-1028.8kJ$

(3) $2NO(g)\rightarrow O_2(g)+N_2(g)$ $\Delta G^o_3=-2\times 87.6kJ=-175.2kJ$

The expression for $\Delta G^o_{rxn}$ will be,

$\Delta G^o_{rxn}=\Delta G^o_{1}+\Delta G^o_{2}+\Delta G^o_{3}$

$\Delta G^o_{rxn}=(72.6kJ)+(-1028.8kJ)+(-175.2kJ)$

$\Delta G^o_{rxn}=-1131.4kJ$

Therefore, the value of $\Delta G^o_{rxn}$ for the reaction is -1131.4 kJ

3 0

3 years ago

If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters, what must the new temperature be to maintain constant pre

Fed [463]

Answer:

893K=620 °C

Explanation:

PV=nRT

P=nRT/V

P= (1 mol* 0.0821* 298K)/15L P=1.63 atm

T=PV/nR

T=(1.63 atm*45 L)/(1 mol*0.0821)

T= 893K=620 °C

7 0

3 years ago