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OLEGan [10]
1 year ago
9

One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w

eight of gas. ​
Chemistry
1 answer:
expeople1 [14]1 year ago
4 0

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

hence, molecular weight of the gas = 32.256g

vapor density = molecular weight/2

= 32.256/2 = 16.128g

hence, vapor density of the gas is 16.128g.

Explanation:

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Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

5 0
2 years ago
The number of organisms that the environment in an ecosystem can handle is called its _____.
Basile [38]
The answer to this question is "carrying capacity." The term "growth rate," refers to how fast a population grows, and the term "population density," refers to the number of organisms located within a specific area. Carrying capacity is correct because is directly addresses the maximum number of organisms that an ecosystem can handle, as opposed to how fast they are growing or how many there currently are.
5 0
2 years ago
Read 2 more answers
This gas can be formed if a hydrogen bomb is detonated.
Levart [38]

Answer:

Helium.

Explanation:

Hydrogen in the bomb is used in the process of detonation. A stream of tritium, an isotope of hydrogen is released and this fissionable material is very unstable thus it turns during the detonation to helium 3. This triggers a series of reactions that produce large amounts of heat to the surrounding environment causing destruction.

7 0
2 years ago
Ten pints of 15 % salt solution are mixed with 15 pints of 10 % salt solution. What is the
sergij07 [2.7K]
Concentration\ rate=\frac{mass\ of\ product}{mass\ of\ substance}*100\%\\\\
15\%=\frac{m_1}{ms_1}*100\%\\ms_1=10pints\\
15\%=\frac{m_1}{10}*100\%\\
\frac{15}{100}=\frac{m_1}{10}\ \ |*10\\
m_1=\frac{150}{100}=1,5pints\\\\
10\%=\frac{m_2}{15}*100\%\\
0,1=\frac{m_2}{15}\ \ |*15\\
1,5=m_2\\
Resulting\ solution:\\
ms=ms_1+ms_2=15+10=25pints\\m=m_1+m_2=1,5+1,5=3pints
\\\frac{m}{ms}*100\%=\frac{3}{25}*100\%=12\%
7 0
2 years ago
N2 molecules absorb ultraviolet light but not visible light. I2 molecules absorb both visible and ultraviolet light. Which of th
Umnica [9.8K]

C. Visible light does not produce transitions between electronic energy levels in the N₂ molecule but does produce transitions in the I₂ molecule.

Explanation:

Abortion of light by molecules will produce electronic transitions from a ground level to a higher level equal to the energy of absorbed light.

In the case of nitrogen (N₂) the allowed electronic transitions are between electronic energy levels with a energy difference equal to the energy of photons of the ultraviolet light. Nitrogen will not absorb from the visible range so it is colorless.

Now iodine (I₂) have the allowed electronic transitions between electronic energy levels with a energy difference equal to the energy of photons of the visible light. As a consequence Iodine vapors have a violet color.

Learn more about:

electronic transitions

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#learnwithBrainly

4 0
2 years ago
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