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tino4ka555 [31]
3 years ago
9

Calculate the volume of the gas when the pressure of the gas is 1.60 atm at a temperature of 298 K. There are 160. mol of gas in

the cylinder. The value for the universal gas constant R is 0.08206 L⋅atm/(mol⋅K) .
Express your answer numerically to four significant figures.
Chemistry
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

2445 L

Explanation:

Given:  

Pressure = 1.60 atm

Temperature = 298 K

Volume = ?

n =  160 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K  

<u>⇒V = 2445.39 L</u>

Answer to four significant digits, Volume = 2445 L

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Balance chemical reaction (dissociation):
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In order to predict the outcome of the reaction, write the molecular, full ionic, and net ionic equations for a mixture of aqueo
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See explanation

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Full molecular equation;

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Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

2NH3(aq) + Ag^+(aq) -------->  [Ag(NH3)2]^+(aq)

When Silver nitrate is mixed with a solution of aqueous ammonia, a white and cloudy solution was observed.

6 0
3 years ago
_Al2O3+heat=_Al+_O2​
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Answer:

6.

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7.

2AI+ 6HCl -> 3H2 + 2AlCl3

Reaction type: Single Replacement

8.

IDK

Reaction type: Double Replacement?

Explanation:

8 0
3 years ago
A sample of nitrogen gas has a volume of 578 mL and a pressure of 124.1 kPa. What volume would the gas occupy at 80.2 kPa if the
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Answer:

V2 = 894.4mL

Explanation:

P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?

Applying Boyle's law

P1V1 = P2V2

Substitute and simplify

124.1*578=80.2*V2

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4 0
4 years ago
Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

75 \times \frac{1}{2} =48.2

34.5 \times \frac{1}{2} =72.3

17.25 \times \frac{1}{2} =96.4

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So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

A=A_{0} \cdot \frac{1}{2^{n}}

Where

A_0  is the initial sample amount  

n = the number of half-lives that pass in a given period of time.

7 0
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