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3 years ago

Which components of atoms has no charge

1 answer:

6 0

The answer is "neutrons." Neutrons have no charge although they are one of the main components to a atom. When the nucleus of the atom forms two particles are formed with it and these are called protons, and neutrons. Remember that a proton has a negative charge, and a neutron has no charge.

Hope this helps!

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What do we do to increase the surface area of solid reactants to high rate of reaction?

motikmotik

The rate of a chemical reaction can be raised by increasing the surface area of a solid reactant. This is done by cutting the substance into small pieces, or by grinding it into a powder.

8 0

3 years ago

A sample of nitrogen gas occupies a volume of 1.40L at 297K with a pressure of 776mmHg. The volume increases to 2.70L and the te

konstantin123 [22]

Hey it isnt letting me submit my answer to your question on the Japanese chart, so imma just submit it here

6 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

4 years ago

If water were a non-polar Molecule

JulijaS [17]

Answer:

Water would not be able to transport nutrients -‐-‐ in plants, or in our bodies -‐-‐ nor to dissolve and transport waste products out of our bodies. ... Cohesiveness, adhesiveness, and surface tension: would decrease because without the +/-‐ polarity, water would not form hydrogen bonds between H20 molecules.

8 0

3 years ago

Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT

marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L = L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

Therefore;

L = 60 + α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0

4 years ago