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likoan [24]
3 years ago
7

2 hydrogen atoms; 3 helium atoms; 5 oxygen atoms; 6 iron atoms

Chemistry
1 answer:
lidiya [134]3 years ago
7 0

Answer:

4chlorine;12thousand ;17zinc;2grams;three 8carbon;15aluminium

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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
Please tell if my answer is right or wrong.
xxTIMURxx [149]
Change the places of 'acts against the motion of an object' and 'causes an object to change speed or direction'
5 0
3 years ago
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
3 years ago
What is the mass of 8.0 x 10^26 UF6 molecules?
gulaghasi [49]
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23

Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles

Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol

Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams




3 0
3 years ago
Which grouping shows a decrease in Intermolecular
Anna11 [10]

Answer:

C) solid, liquid, gas

Explanation:

Which grouping shows a decrease in Intermolecular

Forces of Attraction?

A) gas, liquid, solid B) liquid, solid, gas

C) solid, liquid, gas D) solid, gas, liquid

the further the particles are from each other, the less the intermolecular attraction  they are farthest in a gas, then a liquid, and closest in a solid

7 0
3 years ago
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