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3 years ago

Im hella confused and need some help

Chemistry

1 answer:

tatiyna3 years ago

6 0

An oxide is a molecule that has 2 Oxygen atoms in its empirical formula. Stoichiometry would be used as the ratio of Oxygen to Dioxides is 3:2. So 52.29 would be multiplied by 3/2. The answer is 78.435 Mol of CO2 are formed.

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Convert 1.39 x 10^24 atoms of carbon to moles of carbon

Nimfa-mama [501]

If 1mol ------- is ----------- 6,02×10²³
so x ------- is ----------- 1,39×10²⁴

$x=\frac{1,39*10^{24}*1mol}{6,02*10^{23}}\approx2,31mol$

8 0

3 years ago

Read 2 more answers

V=5, M=10 What is the density?

ch4aika [34]

Answer:

the answer to this question is 2 kilogram/cubic meter

Explanation:

4 0

3 years ago

Someone help me please, I will give you brainliest

Nikitich [7]

Answer:

conserve mass.

Explanation:

3 0

3 years ago

For each balanced reaction, indicate the total number of atoms in the table below.

Afina-wow [57]

(i) Make ammonia number of atoms on reactant side is 8 and on product side is 8. (ii) Separate water number of atoms on reactant side is 6 and on product side is 6. (iii) Combust methane number of atoms on reactant side is 9 and number of atoms on product side is 9.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

(i) Make Ammonia

First we have to write the balanced chemical equation

N₂ + 3H₂ → 2NH₃

Reactant side Product side

N = 2 N = 2

H = 6 H = 6

So total number of atoms on product and reactant side is 8.

(ii) Separate Water

First we have to write the balanced chemical equation

2H₂O → 2H₂ + O₂

Reactant side Product side

H = 4 H = 4

O = 2 O = 2

So total number of atoms on product and reactant side is 6.

(iii) Combust methane

First we have to write the balanced chemical equation

CH₄ + 2O₂ → 2H₂O + CO₂

Reactant side Product side

C = 1 C = 1

H = 4 H = 4

O = 4 O = 4

So total number of atoms on product and reactant side is 9.

Thus from the above conclusion we can say that For each balanced reaction, the total number of atoms are

Reaction Total number of atoms

Reactant side Product side

Make Ammonia 8 8

Separate Water 6 6

Combust Methane 9 9

Learn more about the Balanced Chemical Equation here: brainly.com/question/26694427

#SPJ1

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: For each balanced reaction, indicate the total number of atoms in the table below.

Reaction Total number of atoms

Reactant side Product side

Make Ammonia _________ __________

Separate Water _________ __________

Combust Methane _________ __________

3 0

2 years ago

Magnesium metal burns in air to form a mixture of magnesium oxide (MgO, M = 40.31) and magnesium nitride (Mg3N2, M = 100.95). A

dedylja [7]

Answer:

26.95 %

Explanation:

Air contains the highest percentage of oxygen and nitrogen gases. Magnesium then combines with both of the gases:

$2 Mg (s) + O_2 (g)\rightarrow 2 MgO (s)$

$3 Mg (s) + N_2 (g)\rightarrow Mg_3N_2 (s)$

Firstly, find the total number of moles of magnesium metal:

$n_{Mg} = \frac{1.000 g}{24.305 g/mol} = 0.041144 mol$

Let's say that x mol react in the first reaction and y mol react in the second reaction. This means:

$x + y = 0.041144 mol$

According to stoichiometry, we form:

$n_{MgO} = x mol, n_{Mg_3N_2} = \frac{y}{3} mol$

Multiplying moles by the molar mass of each substance will yield mass. This means we form a total of:

$m_{MgO} = 40.31x g, m_{Mg_3N_2} = \frac{y}{3} 100.95 g =$

The total mass is given, so we have our second equation to solve:

$40.31x + 33.65y = 1.584$

We have two unknowns and two equations, we may then solve:

$x + y = 0.041144$

$40.31x + 33.65y = 1.584$

Express y from the first equation:

$y = 0.041144 - x$

Substitute into the second equation:

$40.31x + 33.65(0.04144 - x) = 1.584$

$40.31x + 1.39446 - 33.65x = 1.584$

$6.66x = 0.18954$

$x = 0.028459$

$y = 0.041144 - x = 0.012685$

Moles of nitride formed:

$n_{Mg_3N_2} = \frac{y}{3} = 0.0042282 mol$

Convert this to mass:

$m_{Mg_3N_2} = 0.0042282 mol\cdot 100.95 g/mol = 0.4268 g$

Find the percentage:

$\omega_{Mg_3N_2} = \frac{0.4268 g}{1.584 g}\cdot 100\% = 26.95 \%$

7 0

3 years ago