What is the molar mass of Aluminum chromate

Which supports the idea that birds and butterflies both have wings but they do not have a common ancestor with wings?

vesna_86 [32]

A. The wings are analogous structures that evolved differently and do not have a similar internal structure.

8 0

2 years ago

he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o

frosja888 [35]

Answer: The number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

Explanation:

We are given:

$K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L$

To calculate the concentration of a substance, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ ......(1)

Concentration of ammonia:

$[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L$

Concentration of ammonium iodide:

$[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L$

For the given chemical reaction:

$NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI][NH_3]}{[NH_4I]}$

Putting values in above equation, we get:

$7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}$

$[HI]=2.53\times 10^{-4}$

Calculating the moles of hydrogen iodide by using equation 1, we get:

$2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}$

Hence, the number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

3 0

2 years ago

What is the limiting reactant when 20g CH4 react with 15g H20?

Art [367]

Answer:

H₂O.

Explanation:

It is clear from the balanced equation:

CH₄ + 2H₂O → CO₂ + 4H₂.

that 1.0 mole of CH₄ reacts with 2.0 moles of H₂O to produce 1.0 mole of CO₂ and 4.0 moles of H₂.

To determine the limiting reactant, we should calculate the no. of moles of (20 g) CH₄ and (15 g) H₂O using the relation:

n = mass/molar mass

no. of moles of CH₄ = mass/molar mass = (20 g)/(16 g/mol) = 1.25 mol.

no. of moles of H₂O = mass/molar mass = (15 g)/(18 g/mol) = 0.833 mol.

from the balanced reaction, 1.0 mole of CH₄ reacts with 2.0 moles of H₂O.

So, from the calculated no. of moles: 0.4167 mole of CH₄ reacts completely with 0.833 mole of H₂O and the remaining of CH₄ will be in excess.

So, the limiting reactant is H₂O.

4 0

2 years ago

After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage

Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have 2 reactions, the formation of the grignard reagent and the formation of the alcohol. The first step then is the calculation of the maximum amount of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the following conversion ratios:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide ( $C_6H_5Br$ )= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and $C_6H_5Br$ = 1:1

$2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg$

$9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br$

The smallest value is the mol of bromobenzene therefore 0.0402 mol of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of formation of the alcohol. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

$4.83~g~\frac{1~mol}{136.14~g}=0.35~mol$

The we have to divide by the coefficient of each reactive in the balance reaction. So:

$\frac{0.35~mol}{1}=0.35$

$\frac{0.0402~mol}{2}=0.0201$

Therefore the limiting reagent would be the phenylmagnesium bromide. Now, the molar ratio between the phenylmagnesium bromide and triphenyl carbinol is 2:1, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to grams of triphenyl carbinol:

Molar mass of triphenyl carbinol= 260.33 g/mol

$0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol$

Finally, we have to divide the obtanied solid by the calculated one:

$Percentage=\frac{5}{5.23}*100=95.6\%$

5 0

3 years ago

Calculate the amount of carbon dioxid gas in 1.505x10^23 molecules of the gas.

Daniel [21]

Explanation:

We need to find the amount of carbon dioxide gas in $1.505\times 10^{23}$ molecules of the gas.

We know that, 1 mole weighs 44 gram of carbon dioxide which contains $6.022\times 10^{23}$ number of molecules. It means that, $6.022\times 10^{23}$ number of molecules present in 44 grams of carbon dioxide molecule. So, $1.505\times 10^{23}$ number of molecules present in :

$=\dfrac{1.0505\times 10^{23}}{6.022\times 10^{23}}\times 44\\\\=7.675\ \text{grams}$

Hence, 7.675 grams of carbon dioxide is present in $1.505\times 10^{23}$ molecules of the gas.

4 0

2 years ago