Pls tell me which answer type it

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

3 years ago

An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula

uysha [10]

Answer:

The answer is: Al2O3

Explanation:

The data they give us is:

0.545 gr Al

0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al

0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

0.02 mol Al / 0.02 = 1 Al

0.03 mol O / 0.02 = 1.5 O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

1Al * 2 = 2Al

1.5O * 2 = 3O

Now the answer is given correctly:

Al2O3

8 0

2 years ago

For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazin

barxatty [35]

Answer:

53.6 g of N₂H₄

Explanation:

The begining is in the reaction:

N₂(g) + 2H₂(g) → N₂H₄(l)

We determine the moles of each reactant:

59.20 g / 28.01 g/mol = 2.11 moles of nitrogen

6.750 g / 2.016 g/mol = 3.35 moles of H₂

1 mol of N₂ react to 2 moles of H₂

Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.

2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine

Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄

Let's convert the moles to mass:

1.67 mol . 32.05 g/mol = 53.6 g

4 0

3 years ago

How many moles of N2 in 57.1 g of N2?

SpyIntel [72]

We are given –

Mass of $\bf N_2$ is 57.1 g and we are asked to find number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\pink{\bf\longrightarrow { Molar \:mass \:of \: N_2:-} }$

$\qquad$ $\bf \twoheadrightarrow 14\times 2$

$\qquad$ $\bf \twoheadrightarrow 28$

$\qquad$ ____________________

Now,Let's calculate the number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\purple{\bf\longrightarrow { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}$

$\qquad$ $\bf \twoheadrightarrow \dfrac{57.1}{28}$

$\qquad$ $\bf \twoheadrightarrow 2.04\: moles$

__________________________________

7 0

2 years ago

Water that flows back to the ocean is called

nordsb [41]

Surface runoff

Explanation:

The water that flows back to the streams and oceans are called surface runoff.

Surface runoff is a component of the water cycle usually composed of water in the liquid form that flows back into oceans that are nearby.

The hydrologic cycle shows the cyclic process by which water passes in nature.
Water passes through different forms, solid, liquid and gases.
Surface runoff is water usually after rainfall that flows rapidly.
They move to the final basin of deposition usually joining up with other water sources.
This can be nearby streams, lakes or oceans.

learn more:

Downcutting a stream brainly.com/question/9259211

#learnwithBrainly

4 0

3 years ago