1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alekssr [168]
3 years ago
5

When 100ml of M HCl is mixed with 100ml of N NaOH, the pH of the resulting solution is

Chemistry
1 answer:
RoseWind [281]3 years ago
4 0

Answer:

Option d. 7

Explanation:

A mixture of a strong base and a strong acid produce a neutral salt and water.

This is the reaction of neutralization:

HCl + NaOH → NaCl + H₂O

NaCl →  Na⁺  +  Cl⁻

Sodium chloride is neutral salt which does not give H⁻ neither OH⁻ to medium, that's why pH is neutral.

Both ions are derivated from a strong acid and base so they do not make hydrolisis. They are a conjugate pair of a weak acid and base. The reactions can not occur:

Cl⁻  +  H₂O ← OH⁻  +  HCl

Na⁺  +  H₃O⁺ ← NaOH  + H₂O

You might be interested in
Draw the product of the reaction CH3CH2CH2CH2C?CCH3+2Br2??
svetlana [45]
I will assume that the sign ? between the C and the CCH3 is a triple bond, and I will represent it by three vertical lines |||


So the reaction is:


<span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->


This is a typical reaction known as halogenation of alkines.


This is an addition reaction, i.e. the alkyne undergoes an addition of the Br2 (and it also happens with Cl2) to the triple bond to form a tetra halide.


.
                                                                                                      Br   Br
</span>                                                                                                        |     |
<span><span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->  CH3 CH2 CH2 CH2 C - C</span> - CH3
                                                                                                        |     |
                                                                                                        Br   Br


</span>
3 0
3 years ago
How many atoms in 5.01 mole of lead?
leva [86]

Answer:

30.17 × 10²³ atoms

Explanation:

Given data:

Number of moles of lead = 5.01 mol

Number of atoms = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

In given question:

1 mole = 6.022 × 10²³ atoms

5.01 mol × 6.022 × 10²³ atoms / 1 mol

30.17 × 10²³ atoms

3 0
3 years ago
What happen when you add heat to solids and liquids?​
sasho [114]
When you add heat to a solid the particles gain energy and start to vibrate faster and faster.
When you add heat to a liquid the particles are given more energy and move faster and faster expanding the liquid.
7 0
2 years ago
Read 2 more answers
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0
3 years ago
Other questions:
  • Calculate the theoretical yield of ammonia produced by the reaction of 100g of H2 gas and 200g of N2 gas
    10·1 answer
  • Quiz. Formula Weight and % Composition
    8·1 answer
  • How many mole are in 72.9g of HCI
    9·1 answer
  • How many moles of H2O are in 12.3L?
    11·1 answer
  • State 3 functions of enzymes in the body
    10·1 answer
  • When ammonium nitrate is added to a suspension of magnesium hydroxide in water, the Mg(OH)2 dissolves. Write a net ionic equatio
    10·1 answer
  • What volume, IN MILLILITERS, of 5.0M HCl solution is required to react with 3.00 g or magnesium metal?
    14·1 answer
  • Please help!! i will give brainiest
    13·1 answer
  • What do calories provude for the human body?
    14·2 answers
  • 1. Shankar and Sameer performed an experiment to differentiate primary, secondary and tertiary amines in a laboratory.Shankar co
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!