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3 years ago

1. An unknown amount of water

Chemistry

1 answer:

m_a_m_a [10]3 years ago

3 0

Answer:

20.93 g

Explanation:

From the question given above, the following data were obtained:

Heat (Q) = 3.5 KJ

Initial temperature (T₁) = 26°C

Final temperature (T₂) = 66°C

Mass (M) =?

Next, we shall convert 3.5 KJ to J. This can be obtained as follow:

1 KJ = 1000 J

Therefore,

3.5 KJ = 3.5 KJ × 1000 J / 1 KJ

3.5 KJ = 3500 J

Next, we shall determine the change in the temperature of the water. This is illustrated:

Initial temperature (T₁) = 26°C

Final temperature (T₂) = 66°C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 66 – 26

ΔT = 40 °C

Finally, we shall determine the mass of the water. This can be obtained as follow:

Heat (Q) = 3500 J

Change in temperature (ΔT) = 40 °C

Specific heat capacity (C) = 4.18 J/gºC

Mass (M) =?

Q = MCΔT

3500 = M × 4.18 × 40

3500 = M × 167.2

Divide both side by 167.2

M = 3500 / 167.2

M = 20.93 g

Therefore, the mass of the water is 20.93 g

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Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans

loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

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Explanation:

We are given:

Molarity of sodium hypochlorite = 0.39 M

$pK_a$ of HClO = 7.50

We know that:

$pK_a=-\log K_a$

$K_a$ of HClO = $10^{-7.50}=3.16\times 10^{-8}$

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant = $3.16\times 10^{-8}$

$K_b$ = Base dissociation constant

Putting values in above equation, we get:

$10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}$

The chemical equation for the reaction of hypochlorite ion with water follows:

$ClO^-+H_2O\rightarrow HClO+OH^-$

Initial: 0.39

At eqllm: 0.39-x x x

The expression of $K_b$ for above equation follows:

$K_b=\frac{[HClO][OH^-]}{[ClO^-]}$

Putting values in above equation, we get:

$3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035$

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.00035M$

Putting values in above equation, we get:

$pOH=-\log (0.00035)=3.6$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-3.6=10.4$

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