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3 years ago

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Chemistry

1 answer:

mihalych1998 [28]3 years ago

7 0

Answer:

67.1%

Explanation:

Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:

Moles Na₂CO₃ - 105.99g/mol-:

6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.

As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:

0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃

And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):

0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.

And percent of NaHCO₃ in the sample is:

10.06g NaHCO₃ / 15g Sample * 100 =

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It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t

irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol

Explanation : To solve this question we can use graham's law;

Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;

So here, Rate of gas 2 / Rate of gas 1 = $\sqrt{(molar mass 1 / molar mass 2)}$

Now, 1.687 = square root [ $\sqrt{(molar mass 1) / (28.01 g/mol N_{2})}$ ]

When we square both the sides we get;

2.845 = (molar mass 1) / (28.01 g/mol N2)

On rearranging, we get,

2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

3 0

3 years ago

The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a

stealth61 [152]

Answer: The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

Explanation:

We are given a function that calculates the amount of sample remaining after 't' years, which is:

$A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}$

For t = 20 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}$

$A_t(t)=288.522g$

Hence, the amount of sample left after 20 years is 288.522 g

For t = 50 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}$

$A_t(t)=144.26g$

Hence, the amount of sample left after 50 years is 144.26 g

6 0

3 years ago

Question 3 (1 point)

pentagon [3]

Full question:

The IUPAC name for CH3CH2C≡CCH3 is:

Answer:

2-pentyne

Explanation:

To name hydrocarbons, you first you have to identify the longest carbon chain. There are 5 carbons in this chain, so we know the name is "pent".

You then have to identify the presence of any double or triple bonds. If double bonds, it is an alkene, if triple bonds, it is an alkyne. In this case there is a triple bond, so we know the hydrocarbon is pentyne.

You then number the chain to give the lowest number to the triple bond. It could either be 4 (countnig carbons from left to right) or 2 (from right to left). Therefore, the answer is 2-pentyne.

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3 years ago

Lead (ll) nitride+ammonium sulfate > lead(ll) sulfate + ammonium nitride

Alenkinab [10]

Answer:

$Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N$

Explanation:

Let's rewrite the given word equation in its chemical balanced equation representation:

1. Lead(II) nitride is represented by lead, Pb, in an oxidation state of 2+, while nitride is a typical nitrogen anion with a state 3-. As a result, the lowest common multiple between 2 and 3 is 6, meaning 2 lead cations are needed to balance 3 nitrogen anions: $Pb_3N_2$ .