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2 years ago

A sample of helium has a volume of 3.49 L at 741 mmHg. What would be the volume if the pressure

Chemistry

1 answer:

Naddika [18.5K]2 years ago

5 0

P1: 741 mmHg

V1: 3.49 L P1 x V1 / P2= (741 mmHg) (3.49 L) / 760 mmHg = 3.40 L

P2: 760 mmHg

V2: ? L

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If a gas has a volume of 1000 ML at a temperature of 23°C and a pressure of 100 mmhg, what is it’s volume under standard conditi

Colt1911 [192]

Answer:

119.7 mL.

Explanation:

From the general law of ideal gases:

PV = nRT.

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

For the same no. of moles of the gas at two different (P, V, and T):

P₁V₁/T₁ = P₂V₂/T₂.

P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.

P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).

∴ V₂ = (P₁V₁T₂)/(T₁P₂) = (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) = 121.4 mL.

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3 years ago

How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)

Masteriza [31]

Answer: The amount of heat required to warm given amount of water is 470.9 kJ

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g$

To calculate the heat absorbed by the water, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ$

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

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3 years ago

A solution has a [oh−] of 1 x 10−12. what is the poh of this solution? 2 7 10 12

Colt1911 [192]

POH = - log [ OH-]

pOH = - log [ 1 x 10⁻¹²]

pOH = 12

8 0

2 years ago

Read 2 more answers

The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days

professor190 [17]

Answer:

6.2 g

Explanation:

In a first-order decay, the formula for the amount remaining after n half-lives is

$N = \frac{N_{0}}{2^{n}}$

where

N₀ and N are the initial and final amounts of the substance

1. Calculate the number of half-lives.

If $t_{\frac{1}{2}} = \text{3.8 da}$

$n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}$

2. Calculate the final mass of the substance.

$\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}$

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3 years ago

Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of

leonid [27]

Volume=250mL
Density=0.71g/ml

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf{:}\implies Mass=Density(Volume)$

$\\ \sf{:}\implies Mass=0.71(250)$

$\\ \sf{:}\implies Mass=177.5g$

6 0

2 years ago