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user100 [1]
3 years ago
7

In a neutralization reaction, an aqueous solution of an Arrhenius acid reacts with an aqueous solution of an Arrhenius base to p

roduce
A. an ether and water
B. an ether and alcohol
C. a salt and water
D. a salt and an alcohol
Chemistry
1 answer:
solmaris [256]3 years ago
5 0

Answer: c. Salt and Water

Explanation:

For example;

When an Arrhenius acid such as; Tetraoxosulphate (VI) acid  (H2SO4) reacts with an Arrhenius base such as Potassium hydroxide (KOH), the products formed in this neutralization reaction is a salt known as ''Potassium Sulphate''  (K2SO4) and ''Water'' (H2O).

H2SO4   +     KOH ------------->   K2SO4      +    H2O

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30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr
Alenkasestr [34]

Answer: Thus concentration of Ca(OH)_2 in mol/dm^3  is 0.011 and in g/dm^3 is 0.814

Explanation:

To calculate the concentration of Ca(OH)_2, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3         1cm^3=0.001dm^3

Putting values in above equation, we get:

1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3

The concentration in g/dm^3 is 0.011mol/dm^3\times 74g/mol=0.814g/dm^3

Thus concentration of Ca(OH)_2 is 0.011mol/dm^3 and 0.814g/dm^3

4 0
3 years ago
Abiotic factors are no living part of an environment. True or False​
oksian1 [2.3K]

Answer:

true

Explanation:

abiotic means nonliving and biotic means living

6 0
3 years ago
Read 2 more answers
20.352 mL of chlorine under a pressure of 680. mm Hg are
Lunna [17]

Answer:

0.01144L or 1.144x10^-2L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 20.352 mL

P1 (initial pressure) = 680mmHg

P2 (final pressure) = 1210mmHg

V2 (final volume) =.?

Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:

P1V1 = P2V2

680 x 20.352 = 1210 x V2

Divide both side by 1210

V2 = (680 x 20.352)/1210

V2 = 11.44mL

Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:

1000mL = 1 L

11.44mL = 11.44/1000 = 0.01144L

Therefore the volume of the container is 0.01144L or 1.144x10^-2L

7 0
3 years ago
Read 2 more answers
How many grams of NaOH needed to completely neutralize 3L of 1.75M HCL
Sedaia [141]

Answer:

209.98 g of NaOH

Explanation:

We are given;

  • Volume of HCl as 3 L
  • Molarity of HCl as 1.75 M

We are required to calculate the mass of NaOH required to completely neutralize the acid given.

First, we write a  balanced equation for the reaction between NaOH and HCl

That is;

NaOH + HCl → NaCl + H₂O

Second, we determine the number of moles of HCl

Number of moles = Molarity × Volume

                             = 1.75 M × 3 L

                             = 5.25 moles

Third, we use the mole ratio to determine the moles of NaOH

From the reaction,

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = Moles of HCl

                          = 5.25 moles

Fourth, we determine the mass of NaOH

Molar mass of NaOH = 39.997 g/mol

Mass of NaOH = 5.25 moles × 39.997 g/mol

                        = 209.98 g

Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl

6 0
3 years ago
Please help me :( Thank you so much ❤️
m_a_m_a [10]

Answer:

G- Gallons-Miles

Explanation

Even though gallons of gas are converted to miles you cannot physically convert gallons of something to miles.

6 0
3 years ago
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