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2 years ago

Explain why by referring to the kinetic particles theory PLZ HELP:(

Chemistry

1 answer:

Olegator [25]2 years ago

5 0

Answer:All matter consists of small particles which are in a continual state of motion

Explanation

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If the temperature of the water an ice cube is placed in is ______________, then the ice cube will melt __________.

Anit [1.1K]

Answer:

I have no clue what is being asked in the second half of the question. Is there some context from which the question is taken?

Explanation:

If the temperature of the water an ice cube is placed in is ___above 0°C__, then the ice cube will melt _unless you remove it??_________.

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2 years ago

In the presence of a base, blue litmus paper will . O turn purple stay blue to turn red

Inessa05 [86]

Answer:

In the presence of a base, blue litmus paper will turn red........

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2 years ago

Read 2 more answers

How many grams of CaCl2 are needed to make 277.8g of a solution that is 31.5% (m/m) in water? Note that mass is not technically

Sveta_85 [38]

Answer: The mass of calcium chloride present in given amount of solution is 87.5 g

Explanation:

We are given:

Mass of solution = 277.8 grams

Also, 31.5 % (m/m) of calcium chloride in water. This means that 31.5 g of calcium chloride is present in 100 g of solution.

To calculate the mass of calcium chloride in the given amount of solution, we use unitary method:

in 100 g of solution, the mass of calcium chloride present is 31.5 g

So, 277.8 g of solution, the mass of calcium chloride present is $\frac{31.5}{100}\times 277.8=87.5g$

Hence, the mass of calcium chloride present in given amount of solution is 87.5 g

8 0

3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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2 years ago

The valance number of M is 3 . what is the formula of its phosphorus

lakkis [162]

The formula for phosphorus is P4

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2 years ago