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astra-53 [7]
3 years ago
8

Match the salt with the acid and base used to form it in a neutralizing reaction. K2SO4

Chemistry
2 answers:
jeyben [28]3 years ago
4 0

for K2SO4 ,

B) KOH and H2SO4

for NaBr,

A) NaOH and HBr

emmasim [6.3K]3 years ago
3 0

Answer:

KOH and H₂SO₄

Explanation:

Neutralization reaction:

It is the reaction in which acid and base react with each other and produce salt and water.

For example:

2KOH + H₂SO₄  →   K₂SO₄ + 2H₂O

1. Potassium hydroxide and sulfuric acid react to produce potassium sulfate salt and water.

2. Potassium hydroxide and phosphoric acid react to produce potassium phosphate and water.

H₃PO₄  + 3KOH  → K₃PO₄ + 3H₂O

3. Phosphoric acid sodium hydroxide react to produce sodium phosphate and water.

H₃PO₄  + 3NaOH   → Na₃PO₄ + 3H₂O

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For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

8 0
3 years ago
A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal
devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0
3 years ago
Substances that are considered acidic are also called alkaline, true or false
V125BC [204]
False
Alkaline is something we use everyday
Acid is something sour eg lemons
7 0
3 years ago
Read 2 more answers
What is the frequency of an electromagnetic wave that has a wavelength of
Korolek [52]

➡ ANSWER

☑ <em><u>C</u></em><em><u>.</u></em><em><u> </u></em><em><u>3.5 105</u></em><em><u> </u></em><em><u>Hz</u></em>

3 0
2 years ago
Hydrogen 3 has a half life of 12.32 years a sample of h-3 weighing 3.02 grams is left for 15.0 years what will the final weight
yawa3891 [41]

Answer:

The final mass of sample is 1.3 g.

Explanation:

Given data:

Half life of H-3 = 12.32 years

Amount left for 15.0 years = 3.02 g

Final amount = ?

Solution:

First all we will calculate the decay constant.

t₁/₂ = ln² /k

t₁/₂ =12.32 years

12.32 y =  ln² /k

k = ln²/12.32 y

k = 0.05626 y⁻¹

Now we will find the original amount:

ln (A°/A) = Kt

ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y

ln (3.02 g/ A) = 0.8439

3.02 g/ A = e⁰°⁸⁴³⁹

3.02 g/ A = 2.33

A = 3.02 g/ 2.33

A = 1.3 g

The final mass of sample is 1.3 g.

8 0
3 years ago
Read 2 more answers
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