All categories

2 years ago

A student wants to determine if a

Chemistry

1 answer:

Sauron [17]2 years ago

7 0

Answer:

The Density of Magnesium.

Explanation:

You might be interested in

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

Does 1 gallon or 100 gallons take longer to boil

Ilya [14]

Answer:

it takes 100 gallons longer to boil then it does 1 gallon

Explanation:

brainliest please??

4 0

2 years ago

Which element or elements are unbalanced in this equation?

Stells [14]

Answer:

Only the H is unbalanced.

Explanation:

There are 4 H's and 2 of everything else

7 0

3 years ago

Read 2 more answers

Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.

Ronch [10]

Answer : The limiting reagent is $O_2$

Solution : Given,

Moles of methane = 2.8 moles

Moles of $O_2$ = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 5 moles of $O_2$ react with $\frac{5}{2}=2.5$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is $O_2$

8 0

3 years ago

Wild animals are not considered a natural resource.

Anvisha [2.4K]

Answer:

False

Explanation:

Wild animals are found in an ecosystem and they are considered to be a natural resource and a biotic factor in the environment. They are living things so they are a natural resource.

4 0

3 years ago