Ask question

Ask question

All categories

3 years ago

12

What mass of water is required to react completely with 157.35 g CO2? (Molar mass of H2O = 18.02 g/mol; molar mass of CO2 = 44.0

1 g/mol

1 answer:

svlad2 [7]3 years ago

6 0

Answer: the answer is 49.05

Explanation:

You might be interested in

An atom of the element iron has an atomic number of 26 and an atomic mass number of 56. if it is neutral, how many protons, neut

allsm [11]

<span> 26 protons, 30 neutrons, 26 electrons</span>

6 0

3 years ago

Read 2 more answers

= 25 X 5 = (use the correct number of sig figs)

Anton [14]

Answer:

1.25 *10^2

Explanation:

25*5 = 125

= 1.25 *10^2

5 0

3 years ago

When a substance is heated ,the kinetic energy of its molecules

Simora [160]

Answer:

When a substance is heated ,the kinetic energy of its molecules also increase.

Explanation:

K.E is directly proportional to T

5 0

3 years ago

PLEASE HELP PICTURE BELOW

seraphim [82]

Answer:

A.

sorry if it's wrong

Explanation:

8 0

3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

4 years ago