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4 years ago

Molar mass of 2(NH4)2SO4

Chemistry

1 answer:

iris [78.8K]4 years ago

4 0

Explanation:

Ammonium sulfate

PubChem CID 6097028

Chemical Safety Laboratory Chemical Safety Summary (LCSS) Datasheet

Molecular Formula (NH4)2SO4 or H8N2O4S

Synonyms AMMONIUM SULFATE 7783-20-2 Diammonium sulfate Mascagnite Sulfuric acid diammonium salt More...

Molecular Weight 132.14 g/mol

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Silver is produced as a byproduct of gold

VashaNatasha [74]

Answer:

Element Properties

atomic number 47

atomic weight 107.868

melting point 960.8 °C (1,861.4 °F)

boiling point 2,212 °C (4,014 °F)

specific gravity 10.5 (20 °C [68 °F])

oxidation states +1, +2, +3

electron configuration [Kr]4d105s1

hope that helps

4 0

3 years ago

A person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m. What is th

xz_007 [3.2K]

Answer:

The efficiency rating of the jack is 0.067.

Explanation:

We have, a person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m.

It is required to find the the efficiency rating of the jack. It is equal to the ratio of output work to the ratio of input work. So,

$\eta=\dfrac{1000\times 0.02}{200\times 1.5}\\\\\eta=0.067$

Thus, the efficiency rating of the jack is 0.067.

6 0

3 years ago

What volume of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid according to

kirill115 [55]

Taking into account the definition of molarity and the stoichiometry of the reaction, the correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

The balanced reaction is:

H₃PO₄ (aq) + 3 NaOH (aq) → Na₃PO₄ (aq) + 3 H₂O(aq)

Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₃PO₄: 1 mole
NaOH: 3 moles
Na₃PO₄: 1 mole
H₂O: 3 moles

Molarity is the number of moles of solute that are dissolved in a given volume.

Molarity is determined by:

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

In this case, 0.42 L of 2.39 M phosphoric acid reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:

$2.39 \frac{moles}{liter}=\frac{number of moles of phosphiric acid}{0.42 liters}$

Solving:

number of moles of phosphiric acid= 2.39 $\frac{moles}{liter}$ * 0.42 liters

number of moles of phosphiric acid= 1.0038 moles ≅ 1 mole

Approaching 1 mole of the amount of phosphoric acid required, then by stoichiometry of the reaction, 3 moles of NaOH are necessary to react with 1 mole of the acid.

Then by definition of molarity and knowing that 6.9 M NaOH is needed, you can calculate the necessary volume amount of NaOH by:

$6.9 \frac{moles}{liter} =\frac{3 moles}{volume}$

Solving:

6.9 $\frac{moles}{liter}$ * volume= 3 moles

$volume=\frac{3 moles}{6.9\frac{moles}{liter} }$

volume= 0.44 L

The correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

Learn more about molarity with this example: <u>brainly.com/question/15406534?referrer=searchResults</u>

5 0

3 years ago

State the law of multiple proportions.

Fofino [41]

Statement that when two elements combine with each other to from more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

8 0

3 years ago

Read 2 more answers

Under laboratory conditions of 25.0 degrees C and 99.5 kPa, what is the maximum number of liters of ammonia that could be produc

Minchanka [31]

Answer:

Volume of ammonia 3 L.

Explanation:

Given data:

Temperature = 25°C ( 25+273= 298 k)

Pressure = 99.5 kpa (99.5/101 = 0.98 atm)

Volume of nitrogen = 1.50 L

Volume of ammonia = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Moles of nitrogen:

PV = nRT

n = PV/RT

n = 0.98 atm × 1.50 L / 0.0821 atm. L/mol. K × 298 K

n = 1.47 /24.5 /mol

n = 0.06 mol

Now we will compare the moles of nitrogen with ammonia.

N₂ : NH₃

1 : 2

0.06 : 2×0.06 = 0.12 mol

Volume of ammonia:

PV = nRT

V = nRT/P

V = 0.12 mol× 0.0821 atm. L/mol. K × 298 K/ 0.98 atm

V = 2.9 atm. L /0.98 atm

V = 3 L

7 0

3 years ago