All categories

2 years ago

How many grams are in 4.5 moles of NaF?

Chemistry

1 answer:

Ilia_Sergeevich [38]2 years ago

6 0

Answer:

3139542g

Explanation:

That's because if you talk about

4.5

moles of sodium fluoride, you would get a really absurd number of grams.

You might be interested in

What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Varvara68 [4.7K]

The dissociation of formic acid is:

$HCOOH \rightleftharpoons HCOO^{-} + H^{+}$

The acid dissociation constant of formic acid, $k_a$ is:

$k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}$

Rearranging the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

pH = 2.75

$pH = -log[H^{+}]$

$[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}$

$pk_a = 3.75$

$k_a = 10^{-3.75} = 1.78\times 10^{-4}$

Substituting the values in the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}$

Hence, the ratio is $\frac{1}{10}$ .

4 0

3 years ago

Manganese(III) oxide is a transition metal compound. The oxidation state of manganese in this compound is _______ , and the chem

Kisachek [45]

Answer:

Hello my Friend! The answer is: Manganese(III) oxide is a transition metal compound. The oxidation state of manganese in this compound is +3 , and the chemical formula of the compound is Mn2O3.

Explanation:

Manganese can have two oxidation states: +2 and +3, but in this case, the "(III)" indicates that in this compound, the state of oxidation is +3.

7 0

3 years ago

Read 2 more answers

Explain atoms contain positively charge protons and negatively charge electrons. why does an ordinary atom have no charge?

tatiyna

An ordinary atom is balanced for instance, lithium has 3 protons and 3 electrons the positively charged proton cancels the negatively charged electron.

THINK math plus 3 and negative 3 = 0 neutral

3 0

3 years ago

Earth is approximately 1.5 x 10^8 km from the sun the. How far is earth from the sun meters

Zielflug [23.3K]

Answer:

${ \tt{1 \: km = 1000m}} \\ { \tt{1.5 \times {10}^{8} km = (\frac{1.5 \times {10}^{8} \times 1000 }{1} )m}} \\ = 1.5 \times {10}^{11} \: metres$

4 0

2 years ago

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac

Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0

3 years ago