All categories

3 years ago

How many moles is 90.93 grams of aluminum?

Chemistry

1 answer:

Rashid [163]3 years ago

6 0

Answer:

C) 3.37 moles .

Explanation:

Hello,

In this case, considering that the atomic mass of aluminum is 27.0 g/mol, the moles in 90.93 g of aluminum are computed via the following dimensional analysis:

$n=90.93 gAl*\frac{1molAl}{27.0gAl} \\\\n=3.37molAl$

Thus, answer is C) 3.37 moles .

Best regards.

You might be interested in

Does gasoline (C8H12) have polar or non-polar molecules?

Archy [21]

Answer:

non polar molecules

Explanation:

Ethyl alcohol will dissolve in water in all proportions because the polar molecules interact with each other freely. In contrast, gasoline is a nonpolar molecule. hope that helped...

3 0

3 years ago

Which kind of magma produces a quiet volcanic eruption

Ilia_Sergeevich [38]

Mafic lava is low viscosity lava that creates quiet eruptions.

3 0

3 years ago

Read 2 more answers

In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is

vova2212 [387]

Answer:

4. the heat of reaction

5 0

3 years ago

A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif

noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = $[A_o]$

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = $[A]=0.06[A_o]$

Half life of the carbon-14 = $t_{1/2}=5730 years$

$k=\frac{0.693}{t_{1/2}}$

$[A]=[A_o]\times e^{-kt}$

$[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}$

$0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}$

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0

3 years ago

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

<u>Answer:</u> The mass of original oxalic acid sample is 6.75 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0

3 years ago