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3 years ago

How many grams of sodium chloride could be produced from 41.3 L of chlorine gas?

Chemistry

1 answer:

babymother [125]3 years ago

6 0

Answer:

The grams of sodium chloride that will be made is 292.5 g. calculation. 2Na + Cl₂ → 2NaCl. step 1: calculate the moles of Na. moles = mass/molar mass. From periodic table the molar mass of Na = 23 g/mol. moles = 115 g/23 g / mol = 5 moles. Step 2 : use the mole ratio to determine the moles of NaCl. Na:NaCl is 2:2 = 1:1 therefore the moles of NaCl is also = 5 moles.

Explanation:

We know sodium is a very active metal and chlorine is also a very reactive non-metal. Usually, metals like to eliminate electrons and non-metals such as halogens like to accept electrons. Sodium readily eliminate its last shell electrons become Na+ cation and chlorine accepts that electron to form Cl-anion.

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6. Look carefully at the following diagram. It does not obey the law of conservation of matter. Explain why it does not follow t

Basile [38]

Answer:

Sort of answer, but it'll lead you in the right direction I hope.

Explanation:

I mean, ofc, the molecules or atoms or whatever the circles are can't be put together that way bc matter says so. It looks like the other diagram is showing water or gas, and if it were turning into the one on the right, then it would usually either settle into water if it was gas or become a single solid object like water. If this is gas, then it doesn't make sense, because gas particles are always separate from each other.

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3 years ago

In the laboratory, a student adds 50.4 mL of water to 16.5 mL of a 0.880 M hydroiodic acid solution. What is the concentration o

Vera_Pavlovna [14]

Answer:

Explanation:

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3 years ago

For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.

Iteru [2.4K]

a) $2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

b) $2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

$2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

$E^0 cell = E^0 (reduction) - E^0 (oxidation)$

= $E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)$

= -0.83 - (-2.71) =1.88V

$2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

$E^0cell$ = $E^0(H^+/H_2) -E^0(Ag^+/Ag)$

$E^0cell$ =-0. - (0.8) =-0.8V

Learn more about the half-reactions here:

https://brainly.in/question/18053421

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2 years ago

What is the atomic number of pt-175 after it undergoes alpha decay

DanielleElmas [232]

Answer:

Explanation:

Alpha decay is one of the most basic forms of radioactive decay. It has an atomic mass of 4 and a proton number of two.

Undergoing an alpha decay will decrease the atomic mass by four and the atomic number by two.

In this particular question, we are particular about the atomic number. The atom has an atomic number of 78. So taking 2 off this gives 76

5 0

3 years ago

What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?

podryga [215]

Answer:

17.934 kg of water

Explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then 1.2 x 10²⁶ molecules will give:

$\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times 10^{23}}$

= 17.934 x 10³ g of water

= 17.934 kg of water

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4 years ago