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krok68 [10]
3 years ago
7

A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new

pressure is 775 mm Hg and the temperature is now 26 °C. There are now ________ mol of gas in the flask.
Chemistry
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

PV=nRT

P= Pressure of the gas = 697 mmHg = 0.92 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

P= Pressure of the gas = 775 mmHg = 1.02 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles

Thus the now the container contains 2.07 moles.

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How many molecules are in 1.75 moles of Caco3?
saw5 [17]

Answer:

1.054 x 10²⁴ molecules

Explanation:

In order to convert moles <em>of any given substance</em> into a number of molecules, we need to use <em>Avogadro's number</em>, which states the number of molecules -or atoms, in the case of elements- present in one mol:

  • In one mol there are 6.023x10²³ molecules.

We now <u>convert 1.75 moles into molecules</u>:

  • 1.75 mol * 6.023x10²³ molecules/mol = 1.05x10²⁴ molecules
5 0
3 years ago
Seawater is typically 3.5% salt and has a density of 1.03 g/mL. How many grams of salt would be needed to prepare enough seawate
Bas_tet [7]

Answer:

Amount of salt needed is around 2.3*10³ g

Explanation:

The salt content in sea water = 3.5 %

This implies that there is 3.5 g salt in 100 g sea water

Density of seawater = 1.03 g/ml

Volume of seawater = volume of tank = 62.5 L = 62500 ml

Therefore, the amount of seawater required is:

=Density*Volume = 1.03g/ml*62500ml = 6.44*10^{4} g

The amount of salt needed for the calculated amount of seawater is:

=\frac{6.44*10^{4}g\ water*3.5g\ salt }{100g\ water} =2254 g =2.3*10^{3} g

8 0
3 years ago
What is the mass of an object with a net force of 10 N accelerating at 2 m/s^2??
xz_007 [3.2K]

Answer:

5kg

Explanation:

Force = Mass x acceleration

F = ma

m = F/a = 10N/2m/s^2

m = 10/2 = 5kg

The standard unit for mass = Kilogram

5 0
3 years ago
How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0
3 years ago
Provided is a diagram showing forces on a box.
Eduardwww [97]

Answer:

5 N left

Explanation:

trust me cccuuhh

4 0
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