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seraphim [82]
3 years ago
11

At 600. K how many moles of gas are in a 1.00 L cylinder at atmospheric pressure?

Chemistry
1 answer:
Anton [14]3 years ago
8 0

Answer:

0.02moles

Explanation:

To answer this question, the general gas law equation is used. The General gas law is:

Pv = nRT

Where; P = standard atmospheric pressure (1 atm)

V = volume (L)

n = number of moles

R = Gas law constant

T = Temperature

For this question; volume = 1.00L, atmospheric pressure (P) = 1 atm, R = 0.0821 L-atm / mol K, T = 600K, n = ?

Therefore; Pv = nRT

n = PV/RT

n = 1 × 1/ 0.0821 × 600

n = 1/49.26

n = 0.0203moles

Hence, there are 0.02 moles of gas.

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MgSO4 x 7H2O:
natka813 [3]

Answer:In regards to writing the formula, we know that magnesium has a charge of 1+ and sulfate, SO4, has a charge of 1-. Sulfate is a polyatomic ion so the charge of the whole thing is (SO4)1-. When you combine them you end up with MgSO4 and the "heptahydrate" is seven water molecules.

Explanation:

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3 years ago
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A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 w
Stels [109]

Answer:

see explanation below

Explanation:

First to all, this is a redox reaction, and the reaction taking place is the following:

2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2

According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:

MhVh = MpVp

h would be the hydrogen peroxide, and p the permanganate.

But like it was stated before, the mole ratio is 5:2 so:

5MhVh = 2MpVp

Replacing moles:

5nh = 2MpVp

Now, we just have to replace the given data:

nh = 2MpVp/5

nh = 2 * 1.68 * 0.0213 / 5

nh = 0.0143 moles

Now to get the mass, we just need the molecular mass of the peroxide:

MM = 2*1 + 2*16 = 34 g/mol

Finally the mass:

m = 0.0143 * 34

m = 0.4862 g

8 0
3 years ago
The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions
musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,  

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,  

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.  

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,  

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,  

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,  

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,  

x as 0.115 and y as 0.165

Now the mass of AgCl will be,  

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,  

278.1 × 0.165 = 45.88 grams.  

3 0
3 years ago
Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.
Jlenok [28]

<u>Answer:</u> The density of gold in g/cm^3 is 19.22g/cm^3

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}

We are given:

Density of gold = 1200lb/ft^3

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into g/cm^3, we get:

\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3

Hence, the density of gold in g/cm^3 is 19.22g/cm^3

6 0
3 years ago
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