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3 years ago

Calculating the volume of 0.05mol/dm3 KOH is required to neutralise 25.0cm3 of 0.0150mol/dm3 HNO3

Chemistry

1 answer:

kvasek [131]3 years ago

8 0

Answer:

You first need to construct a balanced chemical equation to describe the reaction:

KOH + HNO3 ---------> KNO3 + H2O

Work out the no. moles of HNO3 being neutralized:

Moles = Volume x Concentration = (25/1000) x 0.0150 = 0.000375 moles

From the balanced equation the molar ratio of KOH to HNO3 is 1:1 so you also need 0.000375 moles of KOH to neutralise the nitric acid

Now you can work out the volume of KOH required:

Volume = Moles/Concentration = (0.000375)/0.05 = 0.0075 dm^3 = 7.5 cm^3

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3 years ago

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$

$75 \times \frac{1}{2} =48.2$

$34.5 \times \frac{1}{2} =72.3$

$17.25 \times \frac{1}{2} =96.4$

$8.625\times \frac{1}{2} =120.5$

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

$A=A_{0} \cdot \frac{1}{2^{n}}$

Where

$A_0$ is the initial sample amount

n = the number of half-lives that pass in a given period of time.

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3 years ago

When the pressure that a gas exerts on a sealed container changes from 1100 bar to 75.5 bar, the temperature changes from k to 2

Serjik [45]

Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
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$\frac{P1}{T1} = \frac{P2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
$\frac{1100 bar}{T} = \frac{75.5 bar}{298 K}$
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3 years ago