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prohojiy [21]
4 years ago
5

In an experiment, an unknown gas effuses at one-half the speed of oxygen gas, which has a molar mass of 32 g/mol. which might be

the unknown gas?
Chemistry
1 answer:
goblinko [34]4 years ago
5 0
If we assume the rate of diffusion of oxygen is 1 , then that of the unknown gas is 1/2.
From the Grahams law of diffusion;
R1/R2= √mm2/√mm1
 1/0.5 = √mm2/√32
   4 = mm2/32
mm2 = 128
Therefore the molecular mass of the unknown gas i 128 g/mol
I therefore think the gas is Hydrogen iodide
since, H=1, I= 127 , thus HI = 128 g/mol
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How many liters of N2 gas are in 2.4 moles at STP?
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Which element(s) are not balanced in this equation?
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Check every element at both sides

Element      left side      right side   conclusion

Fe                2                1                not balanced
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6 0
4 years ago
Read 2 more answers
A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go
4vir4ik [10]

Answer :  The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of NH_4Cl = 26.5 g

Mass of NaOH = 10 g

Molar mass of NH_4Cl = 53.5 g/mole

Molar mass of NaOH = 40 g/mole

First we have to calculate the moles of NH_4Cl and NaOH.

\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

NH_4Cl+NaOH\rightarrow NH_4OH+NaCl

From the balanced reaction we conclude that

As, 1 mole of NaOH react with 1 mole of NH_4Cl

So, 0.25 moles of NaOH react with 0.25 moles of NH_4Cl

From this we conclude that, NH_4Cl is an excess reagent because the given moles are greater than the required moles and NaOH is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of NH_4Cl.

\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl

\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g

Therefore, the amount of reactant left in excess is 13.1075 grams.

5 0
3 years ago
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