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3 years ago

What mass of oxygen contains the same number of molecules as 42g of nitrogen?

Chemistry

1 answer:

lakkis [162]3 years ago

3 0

Answer:

Given: 42 g of N2

Solve for O2 mass that contains the same number of molecules to 42 g of N2.

Solve for the number of moles in 42 g of N2

1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles

Solve for mass of 1 mole of oxygen

1 mole of O2 = 16 g * 2 = 32 g per mole

Solve for the mass of 1.5 moles of oxygen

mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles

mass of 1.5 moles of O2 = 48 g

So 48 g of O2 contains the same number of molecules as 42 g of N2

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"I think my bike is making a scraping noise because the bearings are not lubricated."

GuDViN [60]

Answer:

Explanation:

"I think my bike is making a scraping noise because the bearings are not lubricated."

4 0

2 years ago

A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The ball

yanalaym [24]

Answer:

4.33 L

Explanation:

Step 1: Given data

Initial volume of the balloon (V₁): 3.00 L

Initial pressure of the balloon (P₁): 765 torr

Final volume of the balloon (V₂): ?

Final pressure of the balloon (P₂): 530 torr

Step 2: Calculate the final volume of the balloon

If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.

$P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L$

7 0

2 years ago

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.

3 0

2 years ago

What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52

Andru [333]

Answer:

101,37°C

Explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m (1)

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = 3,27moles of ethlyene glycol

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

X = 101,37°C

I hope it helps!

7 0

3 years ago

I need help with all of it pls and quick

Mazyrski [523]

With what are you need help?

5 0

2 years ago