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3 years ago

5

Which of the following gases is NOT found in Earth’s atmosphere?

2 answers:

netineya [11]3 years ago

5 0

D. all of the above are in the atmosphere

Andrew [12]3 years ago

4 0

The answer is D. All of the above.

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Iodine chloride, ICl, can be made by the following reaction between iodine, I2, potassium iodate, KIO3, and hydrochloric acid.

Vadim26 [7]

The balanced chemical reaction is:

<span>2 I2 + KIO3 + 6 HCl ---------> 5 ICl + KCl + 3 H2O
</span>
We are given the amount of the product to be produced from the reaction. This will be the starting point of our calculations.

28.6 g ICl (1 mol / 162.35 g ICl ) ( 2 mol I2 / 5 mol ICl ) ( 253.81 g I2 / 1 mol I2 ) = 17.88 g I2

6 0

3 years ago

Read 2 more answers

Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,

7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 ) $N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$ ; Kc = 1.54e - 31

2) $N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)$ ; Kc = 2.16e - 24

upon reversing ( 2 ) equation

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$ Kc = 1/2.16e - 24

now adding 1 and reversed equation (2)

$N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$

we get ,

$N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)$ Kc = 1.54e-31 × 1/2.61e - 24

equilibrium constant of equation (3) is -

Kc = 1.54e - 31 / 2.61e - 24

3 0

3 years ago

A bond exists between A and B in the compound AB. A and B are sharing electrons. The type of bond is

Jlenok [28]

Answer: Covalent Bond.

6 0

3 years ago

Use the following balanced reaction to solve:

Naily [24]

Answer: 60.7 g of $PH_3$ will be formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}$

$\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles$

The balanced chemical reaction is

$P_4(s)+6H_2(g)\rightarrow 4PH_3(g)$

$H_2$ is the limiting reagent as it limits the formation of product and $P_4$ is the excess reagent.

According to stoichiometry :

6 moles of $H_2$ produce = 4 moles of $PH_3$

Thus 2.68 moles of $H_2$ will produce= $\frac{4}{6}\times 2.68=1.79moles$ of $PH_3$

Mass of $PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g$

Thus 60.7 g of $PH_3$ will be formed by reactiong 60 L of hydrogen gas with an excess of $P_4$

3 0

3 years ago

You have 5 cars and you double that amount.How much cars do you have.

Nostrana [21]

Answer:

10 cars

Explanation:

the double of 5 is 10

5+5=10

5 0

3 years ago