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Vladimir [108]
3 years ago
8

Tim wants to store 0.122 moles of helium gas at 203 kPa and 401 K. Which is the volume of the helium gas needed for storage?

Chemistry
1 answer:
algol133 years ago
3 0

 The  volume  of Helium gas needed for storage  is 2.00 L (answer C)


 <u><em> calculation</em></u>

 The volume of Helium is calculated using ideal gas equation

That is Pv =nRT

where;

 P( pressure) = 203 KPa

V(volume)=?

n(number  of moles) =  0.122 moles

R(gas constant) = 8.314  L.Kpa/mol.K

T(temperature)= 401 K

make V the  subject of the formula  by diving both side  by P

V=nRT/p

V={[0.122  moles x 8.314 L. KPa/mol.K  x 401 K]  / 203 KPa} = 2.00 L


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Answer:

5Atm

Explanation:

I just guess and it’s right

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Liquid 1 has a density of 2.0 g/ml, Liquid 2
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Liquid 2 because the lower the density the more it floats and the higher the density the more it sinks. The order from top to bottom is liquid 2, liquid 3, liquid 1

8 0
3 years ago
What is the solution to the problem to the correct number of significant figures (102,900/12)+(170•1.27)
GarryVolchara [31]

As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.


_______________________________


102900/12 = 8575


170 × 1.27 = 215.9


∴ (102,900 ÷ 12) + (170 × 1.27) =  8575 + 215.9


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8 0
3 years ago
Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is CO3^{-2}, and the ion nitrate is NO3^{-}.

Sodium is in group 1, so it must lose one electron to be stable, and be the cation Na^{+}. Silver has only one electron too, so the cation will be Ag^{+}.

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0
2 years ago
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they are all catalysts


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