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3 years ago

How many grams are there in 1.70 moles of KMnO4?

Chemistry

2 answers:

topjm [15]3 years ago

8 0

Answer:

269g

Explanation:

hope this helps you

Rashid [163]3 years ago

6 0

Answer: 269 g
Explanation: since we want moles>grams you would multiply the given by the molar mass of the element divided by one.So 1.70*158/1=269 grams

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SSA PRACTICE QUESTION

Leto [7]

Answer:

The correct answer is - so that other scientists can replicate the experiment and make sure the results are correct or to check accuracy.

Explanation:

The data and procudres are required to be recorded or noted correctly so the experiment and research can be replicate and tested for the accuracy of the experiment by other scientist and researcher.

Mandy also need to note procedure, data, variables and other data correctly for the testing the accuracy and replication of the experiment. The replication is essential to check if every thing is correct and result are error free.

4 0

3 years ago

a chemist produces 460 mL of oxygen gas at -43 C and constant pressure. to what celsius temperature must the oxygen be warmed in

Fudgin [204]

Answer:

27 °C

Explanation:

Applying,

V/T = V'/T'................. Equation 2

Where V = Initial volume of oxygen, T = Initial temperature, V' = Final volume of oxygen, T' = Final temperature.

Make T' the subject of the equation

T' = V'T/V................ Equation 2

Form the question,

Given: V' = 600 mL, V = 460 mL, T = -43°C = (-43+273) = 230K

Substitute these values into equation 2

T' = (600×230)/460

T' = 300 K

T' = (300-273) °C

T' = 27 °C

3 0

3 years ago

Three of the primary components of air are

Darya [45]

Answer:

125.681 torr

Explanation:

The formula to calculate the partial pressure of oxygen:

P(O2) = Pcommon - PCO2 - PN2

1 atm = 760 torr

Therefore you need to find the partial pressure of oxygen:

P(O2) = 760 torr -0.285 torr - 634.034 torr =130.013 torr

- Hope that helped!

8 0

4 years ago

Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

Marat540 [252]

Answer : The partial pressure of helium is, $1.815\times 10^4KPa$

Solution : Given,

Molar mass of $O_2$ = 32 g/mole

Molar mass of helium = 4 g/mole

Molar mass of $N_2$ = 28 g/mole

Total pressure of gas = $2.07\times 10^4KPa$

As we are given gases in percent, that means 10 g of oxygen gas, 50 g of helium gas and 40 g of nitrogen gas present in 100 g of mixture.

First we have to calculate the moles of oxygen, helium and nitrogen gas.

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{10g}{32g/mole}=0.3125moles$

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{50g}{4g/mole}=12.5moles$

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{40g}{28g/mole}=1.428moles$

Now we have to calculate the total number of moles of gas mixture.

$\text{Total number of moles of gas}=\text{Moles of oxygen gas}+\text{Mole of helium gas}+\text{Moles of nitrogen gas}$

$\text{Total number of moles of gas}=0.3125+12.5+1.428=14.24moles$

Now we have to calculate the moles fraction of helium gas.

$\text{Mole fraction of He gas}=\frac{\text{Moles of He gas}}{\text{Total number of moles of gas}}=\frac{12.5}{14.25}=0.877$

Now we have to calculate the partial pressure of helium.

$p_{He}=X_{He}\times P_T$

where,

$p_{He}$ = partial pressure of helium

$P_T$ = total pressure

$X_{He}$ = mole fraction of helium

Now put all the given values in this formula, we get

$p_{He}=(0.877)\times (2.07\times 10^4KPa)=1.815\times 10^4KPa$

Therefore, the partial pressure of helium is, $1.815\times 10^4KPa$

8 0

3 years ago

What is the oxidation number of carbon in the compound carbon dioxide, CO2?

Anna35 [415]

Answer:

the oxidation number is 4

7 0

3 years ago