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3 years ago

Why should you always condition a buret before running a titration?.

Chemistry

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

Conditioning two or three times will insure that the concentration of titrant is not changed by a stray drop of water.

Explanation:

"Check the tip of the buret for an air bubble. To remove an air bubble, whack the side of the buret tip while solution is flowing".

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What is the heat transfer mechanism responsible for the daily wind cycle? A. Heat transfer through conduction B. Heat transfer t

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Answer:

the answer is c on plato!!!!!

Explanation:

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4 years ago

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CO2(g) + H2(g)

kondaur [170]

Answer:

Hydrogen

Explanation:

This may be wrong but I can't really tell because of how you have written the question

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3 years ago

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A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sampl

diamong [38]

<u>Answer:</u>

<u>For A:</u> The average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u> The rate of effusion of $SO_2$ gas is $1.006\times 10^{-3}mol/hr$

<u>Explanation:</u>

The average molecular speed of the gas is calculated by using the formula:

$V_{gas}=\sqrt{\frac{8RT}{\pi M}}$

$V_{gas}\propto \sqrt{\frac{1}{M}}$

where, M is the molar mass of gas

Forming an equation for the two gases:

$\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}$ .....(1)

Given values:

$V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol$

Plugging values in equation 1:

$\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s$

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

$Rate\propto \frac{1}{\sqrt{M}}$

Where, M is the molar mass of the gas

Forming an equation for the two gases:

$\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}$ .....(2)

Given values:

$Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol$

Plugging values in equation 2:

$\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr$

Hence, the rate of effusion of $SO_2$ gas is $1.006\times 10^{-3}mol/hr$

8 0

3 years ago

Iron is extracted from iron oxide in the blast furnace. Calculate the maximum theoretical mass of iron that can be made from 100

hichkok12 [17]

<h3>Answer:</h3>

69.918 g

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of iron oxide as 100 g

We are supposed to determine the maximum theoretical yield of Iron from the blast furnace;

The equation for the reaction in the blast furnace that extracts Iron from iron oxide is given by;

Fe₂O₃ + 3CO → 2Fe + 3CO₂

We can first determine moles of Iron oxide;

Moles = Mass ÷ Molar mass

Molar mass of Fe₂O₃ = 159.69 g/mol

Therefore;

Moles of Fe₂O₃ = 100 g ÷ 159.69 g/mol

= 0.626 moles

Then we determine moles of Iron produced

From the equation;

1 mole of Fe₂O₃ reacts to produce 2 moles of Fe

Therefore;

Moles of Fe = Moles of Fe₂O₃ × 2

= 0.626 moles × 2

= 1.252 moles

Maximum theoretical mass of Iron that can be obtained

Mass = Moles × molar mass

Molar mass of Fe = 55.845 g/mol

Therefore;

Mass of Fe = 1.252 moles × 55.845 g/mol

= 69.918 g

Therefore, the maximum theoretical mass of Iron metal obtained is 69.918 g

7 0

4 years ago

How many grams do 6.7×1023 atoms of argon weigh?

boyakko [2]

Answer:

44.4g

Explanation:

First we must find the moles of argon.

To do this, divide the number of atoms by Avogadro's number (6.022 ×10^23)

n= (6.7 × 10^23)/(6.022 × 10^23)

n=1.11258718034

Now to find mass, we complete the calculation

m= n × MM

where

m is mass

n is moles

MM is molecular mass

m= 1.11258718034 × 39.948

m= 44.4456326802

m= 44.4g

4 0

4 years ago