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Nonamiya [84]
3 years ago
9

A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. w

hat volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Chemistry
1 answer:
o-na [289]3 years ago
4 0

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

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A
BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of SO_2 molecules in 9.87 moles  SO_2 is 59.43 molecules.

The moles of Fe in 1.40 x 10^{22} atoms of Fe is 0.23 x 10^{-1}

The moles of C_2H_6O in 2.30x10^{24} molecules of C_2H_6O is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × 10^{23} x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of SO_2 molecules in 9.87 moles of SO_2:

6.02214076 × 10^{23} x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x 10^{22} atoms of Fe:

1.40 x 10^{22} ÷ 6.02214076 × 10^{23}

0.2324754694 x 10^{-1} moles.

0.23 x 10^{-1} moles.

D. The moles of C_2H_6O in 2.30x10^{24} molecules of C_2H_6O:

2.30x10^{24} ÷ 6.02214076 × 10^{23}

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0
2 years ago
2. Write the chemical equations for the neutralization reactions that occurred when HCL and NaOH were added to the buffer soluti
lutik1710 [3]

Answer:

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Explanation:

A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.

A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equations for the neutralizations that occurred upon addition of HCl or NaOH are;

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

5 0
2 years ago
What two motions combine to produce orbit?
Alex73 [517]
<span>Forward & falling. Hope this helps!</span>
6 0
3 years ago
Which of the following metals is in a liquid state at room temperature?
SpyIntel [72]
The answer is c. hg (mercury)
5 0
3 years ago
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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