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yarga [219]
4 years ago
6

A sample of oxygen has a volume of 128 milliliter at a pressure of 500.0 mmhg calculate the volume this sample would occupy at s

tandard pessure
Chemistry
1 answer:
leva [86]4 years ago
4 0

Answer:

84.2mL

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 128mL

Initial pressure (P1) = 500mmHg

Final pressure (P2) = stp = 760mmHg

Final volume (V2) =..?

Thus, the new volume of the gas can be obtained by using the Boyle's law equation as follow:

P1V1 = P2V2

500 x 128 = 760 x V2

Divide both side by 760

V2 = (500 x 128)/760

V2 = 84.2mL

Therefore, the new volume of the gas is 84.2mL

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Ksp(Sr(OH)₂) = [Sr²⁺]·[OH⁻]².<span>
[</span>Sr²⁺] = s.<span>
[</span>OH⁻] = [Sr²⁺] = 2s<span>
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7 0
4 years ago
What products are always made when any carbon-based fuel
Elodia [21]
Answer:
d) carbon dioxide and water
8 0
3 years ago
What is the name of Bel on the periodic table
rewona [7]

Answer:

Nobelium or Beryllium

4 0
3 years ago
The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad
Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}

2\times x+2\times 0.564x=566

2x+1.128x=566

3.128x=566

x=180.9\approx 181pm

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

4 0
3 years ago
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
3 years ago
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