Which of the following will neutralize an acid?

Whats between -999 and 1000

Doss [256]

Hello!

Numbers from -998 and 999 are in between -999 and 1000.

I hope this was helpful! c:

7 0

3 years ago

12. How much mass is in a 3.25-mole sample of NH 4 OH? A. 10.8 g B. 34.0 g C. 35.1 g D. 114 g

Galina-37 [17]

Answer:

D. 114 g

Explanation:

NH4OH molecular weight. Molar mass of NH4OH = 35.0458 g/mol This compound is also known as Ammonium Hydroxide.
Convert grams NH4OH to moles or moles NH4OH to grams. Molecular weight calculation: 14.0067 + 1.00794*4 + 15.9994 + 1.00794.

8 0

3 years ago

1. A gas having the following composition is burnt under a boiler with 50% excess air.

jeka94

The composition of the stack gas are :

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of $CH_4$ : 70% of 100 g = 70 g

Mass of $C_3H_8$ : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of $CH_4$ : $\frac{70 g}{16.04 g/mol}$ = 4.3 mole

Number of moles of $C_3H_8$ : $\frac{15 g}{144.1 g/mol}$ = 0.10 mole

Mass of CO : $\frac{15 g}{28.01 g/mol}$ = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of $CH_4$ : $\frac{4.3}{4.935}$ = 0.8713

Mole fraction of $C_3H_8$ : $\frac{0.10}{4.935}$ = 0.0202

Mole fraction of CO : $\frac{0.53}{4.935}$ = 0.107

Hence, composition of the stack gas are:

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

Learn more about mole fraction here:

brainly.com/question/13135950

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8 0

2 years ago

Acetic acid, CH3COOH, can be produced by bubbling oxygen gas into acetaldehyde, CH3CHO, in the presence of

slamgirl [31]

Explanation:

The balanced equation for the reaction is given as;

2CH3CHO + O2 → 2CH3COOH

If 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel, (a)

how many grams of acetic acid will be produced?

First thing's first, we have to find he limiting reactant. This is done by comparing the number of moles of the reactants.

From the equation of the reaction;

2 mol of CH3CHO reacts with 1 mol of O2

From the masses given;

Number of moles = Mass / Molar mass

CH3CHO;

Number of moles = 20 / 44.0526 = 0.454 mol

O2;

Number of moles = 10 / 32 = 0.3125 mol

The limiting reactant is CH3CHO because O2 would be in excess.

Back to the question;

2 mol of CH3CHO produces 2 mol of CH3COOH

0.454 mol would produce x

Solving for x;

x = 0.454 * 2 / 2 = 0.454 mol

Converting to mass;

Mass = number of moles* Molar mass

Mass = 0.454 mol * 60.052 g/mol = 27.26 grams

(b) how many grams of the excess reactant remain after the reaction is

complete

The excess reactant is O2

Number of moles left = Initial Number of moles - Number of moles that reacted

Number of moles left = 0.3125 mol - (0.454 mol / 2)

Number of moles left = 0.0855 mol

Converting to mass;

Mass = 0.0855 mol * 32 g/mol = 2.736 grams

6 0

3 years ago

How many protons are in the nucleus of this atom?<br> What is the mass number of this atom?

Anettt [7]

Ummmmmmmmmmmmmmmmmmmmmmmmm, 21

3 0

3 years ago