. Which of the following factors contributes to the increase in lonization energy from left to right across a period?

balandron [24]

$\huge\fbox{Answer ☘}$

It is given in the question that 200.0 grams of sodium hydroxide reacts with excess of sulphuric acid.

The given reaction is shown below

$\bold\blue{2NAOH + H _{2}SO _{4} \: - > 2H _{2} O+ NA _{2}SO_{4} }\\$

In this reaction, two mole of sodium hydroxide react with one mole of sulphuric acid to give two mole of water and one mole of sodium sulfate.

To determine the mass of sodium sulfate, first calculate the moles of sodium hydroxide as the value of mass is provided.

The molecular weight of NaOH is 40 g/mol.

The formula is shown below.

$\pink{n = \frac{m}{ M}}\\$

Where,

✫ n is the number of moles

✫ m is the mass

✫ M is the molecular weight

To calculate the number of moles, substitute the values in the above equation.

$\pink{n = \frac{200.0}{40g \: mol {}^{ - 1} } } \\ \\\pink{ ⇢n = 5}$

In the equation it is given that 2 mole of sodium hydroxide gives 1 mole of sodium sulphate.

So, to determine the actual mole of sodium sulfate divide the calculated moles of NaOH by 2.

$\pink{⇢ \frac{5}{2}} \\ \\ \pink{⇢ 2.5 \: mol}$

So 2.5 mol sodium sulfate is formed from 5 mol NaOH.

The molecular weight of sodium sulfate is 142.1 g/mol.

To calculate the mass of sodium sulfate, substitute the values in the formula.

$\pink{⇢2.5 = \frac{m}{142.1g \: mol {}^{ - 1} } } \\ \\\pink{ ⇢ m = 2.5 \times 142.1 }\\ \\ \pink{⇢ m = 355.25}$

Therefore, the mass of sodium sulfate formed when we start with 200.0 grams of sodium hydroxide and you have excess sulphuric acid is 355.25g.

hope helpful~

5 0

2 years ago

A gold nugget has a volume of 4.35cm and a mass of 84.96g. What is its density?

Minchanka [31]

Density equal mass/volume
so you would have 84.96/4.35= 19.5310345

i hope i remembered this right

3 0

4 years ago

Read 2 more answers

Using the following thermochemical data: 2Y(s) + 6HF(g) → 2YF3(s) + 3H2(g) ΔH° = –1811.0 kJ/mol 2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2

Luba_88 [7]

Answer:

ΔH° = 182.4 kJ/mol

Explanation:

The ΔH wanted is for the reaction :

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)

This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.

Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:

2YF3(s) + 3H2(g) → 2Y(s) + 6HF(g) ΔH° = 1811.0 kJ/mol (change sign)

dividing by two :

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol

dividing this one by two,

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

Now adding 1 and 2

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

________________________________________________________

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g). ΔH° = 905.5 + (-723.1) kJ/mol

ΔH° = 182.4 kJ/mol

Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.

8 0

4 years ago

What is milliliter the same as?

Klio2033 [76]

Answer:

1 cubic centimeter

Explanation:

8 0

3 years ago

Read 2 more answers

2 ways to incress the density of water

AlexFokin [52]

Low temperature in water and an increase in salinity.

3 0

3 years ago