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3 years ago

What is the mass, in grams, of 0.125 L of CO2 at STP? Question 4 options: 2.80 g 181 g 0.246 g 4.11 g

1 answer:

7 0

1) Use the fact that 1 mol of gas at STP occupies 22.4 liter

=> 1 mol / 22.4 l = x / 0.125 l => x = 0.125 l * 1 mol / 22.4 l = 0.00558 mol

2) Now use the molar mass of the gas

molar mass of CO2 ≈ 44 g / mol

Formula: molar mass = mass in grams / number of moles =>

mass in grams = molar mass * number of moles = 44 g/mol * 0.00558 moles

mass = 0.246 g

Answer: 0.246 g

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2 years ago

What is the name of the product isotope formed when Thorium-234 decays by beta decay?

defon

Answer:

protactinium-234,

Explanation:

This is what happens when thorium-234 releases a W- boson, which then decays to an electron and an electron antineutrino.

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3 years ago

As the pressure on the surface of a liquid decreases, the temperature at which the liquid will boil

ankoles [38]

Answer: option (1) decreases.

Explanation:

May be you have experienced that: when you go to the beach, where the atmposhpere pressure is greater than the atmosphere pressure in places that are at higher altitudes, the water takes longer to boil. That is because the boiling temperature is greater, and you need more total heat (more time) to permit the liquid to reach that temperature.

The reason why that happens is because substances boil when the vapor pressure (the pressure of the particles of vapor over the liquid) equals the atmosphere pressure. So, when the atmposhere pressure increases, the temperature at which the vapor pressure reaches the atmosphere pressure also increases, and when the atmosphere pressure decreases, the temperature at which the vapor pressure reaches the atmosphere pressure decreases.

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3 years ago

Read 2 more answers

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°

miskamm [114]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H_1=157kJ$ ( × 4)

(2) $P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)$ $\Delta H_2=-1207kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

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3 years ago

The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con

Aloiza [94]

Use the formula for second order reaction:

$\frac{1}{C} = \frac{1}{C_0} + kt$

C = concentration at time t
C0 = initial conc.
k = rate constant
t = time

1st equation : $\frac{1}{0.56} = \frac{1}{C_0} + 115k$

2nd Equation: $\frac{1}{0.033} = \frac{1}{C_0} + 735k$

Find $\frac{1}{C_0}$ from 1st equation and put it in 2nd equation:

$\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k$

$\frac{1}{0.033} - \frac{1}{0.56} = 620k$

k = 0.046

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3 years ago