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2 years ago

You are given a pure sample of a colourless gas in a test tube. It is

Chemistry

1 answer:

Snowcat [4.5K]2 years ago

3 0

Answer:

Explanation:

If the gas pops when the top of the test tube is placed close to a flame it is hydrogen,

If it re-ignites a glowing splint it's oxygen.

Otherwise the gas is nitrogen.

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If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?

solmaris [256]

Answer:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂ = 4.9kg = 4900g

Unknown:

Number of molecules = ?

Solution:

To find the number of molecules, we need to find the number of moles first.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CO₂ = 12 + 2(16) = 44g/mol

Number of moles = $\frac{4900}{44}$ = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

1 mole = 6.02 x 10²³molecules

111.36 moles = 111.36 x 6.02 x 10²³molecules = 6.7 x 10²⁶molecules

5 0

4 years ago

The chemical name for the compound formula NaCl is?

Mandarinka [93]

Answer:

The chemical name for the compound formula NaCl is sodium chloride.

7 0

3 years ago

Read 2 more answers

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

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3 years ago

Lakes can disappear due to *

lesya [120]

Answer:

Drought, resulting in a 65% reduction in water levels.

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3 years ago

HI, I need help my my edg. Can you help me? I will give crown.<br> :D

ahrayia [7]

Answer:

Most liking the puck will go flying because of the force of the hockey stick.

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3 years ago