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DENIUS [597]
3 years ago
11

If you want to produce 6.2 moles nacl, how many moles of na will you need if you have excess cl2

Chemistry
1 answer:
irinina [24]3 years ago
3 0

Answer:

6.2moles of Na are needed.

Explanation:

First, we need to generate a balanced equation for the reaction between sodium and chlorine to produce sodium chloride. This is illustrated below:

2Na + Cl2 —> 2NaCl

From the equation,

2moles of Na produced 2moles of NaCl.

Therefore Xmol of Na will produce 6.2moles of NaCl i.e

Xmol of NaCl = (2 x 6.2)/2 = 6.2moles

Therefore, 6.2moles of Na are needed to produce 6.2moles of NaCl

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Why do larger gases such as Neon produce more color bands (line spectra) than smaller gases like Hydrogen?
LekaFEV [45]
Larger gases produces more spectral lines than the smaller gases because they have more orbitals in their atoms.
Hydrogen has only one orbital in which an electron orbits. At the excited state, that is, when the electron gains energy, the number of energy level it can transcend is very few. For larger elements, they have more orbitals and when excited, they can move from the ground state to other energy levels at which they produce various unique spectral lines.
7 0
3 years ago
How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)
user100 [1]

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

4 0
3 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi
Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL

3 0
3 years ago
Which of the following would most likely happen if water did not form hydrogen bonds?
Oksanka [162]
Water would not expand when it freezes.<span> </span>
8 0
3 years ago
Read 2 more answers
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