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3 years ago

Any substance which yields a hydrogen ion when placed in a water solution is call an——-

Chemistry

1 answer:

lidiya [134]3 years ago

6 0

it is A Acid

Explanation : Any substance that yields a hydrogen ion when placed in a water solution is called an acid. According to Arrhenius theory of Acids and Bases, Acids are those substances which on dissociation in solution generates ions. Whereas a Base is a substance that dissociates in the solution to produce ions.

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

3 years ago

If you start a chemical reaction with 5 atoms of hydrogen, how many hydrogen atoms will you have when it is finished? Group of a

mote1985 [20]

Answer:

5 atoms

Explanation:

According to the law of conservation of mass, "matter is neither created nor destroyed in the cause of a chemical reaction".

We finish with what we start with in a chemical reaction. Although new species might form, the number of atoms on both sides of the expression will still be maintained.

All chemical reactions obey this law of conservation.

8 0

3 years ago

Cirrhosis is a disease that affects which of the listed organs?

Inessa [10]

Cirrhosis is a disease that affects the liver

4 0

3 years ago

Read 2 more answers

You can purchase nitric acid in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. Describe exactly

Andrej [43]

The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.

6 0

3 years ago

Read 2 more answers

g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empir

sammy [17]

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

7 0

3 years ago