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zheka24 [161]
2 years ago
8

How many grams of sodium hydroxide (NaOH) do you need to get rid of 3000 grams of CO2? Don't worry about sig figs for this quest

ion.
Chemistry
1 answer:
Slav-nsk [51]2 years ago
8 0

Answer:

It would require 80 g of NaOH .

Explanation:

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This is my chemistry worksheet. It's a new topic my teacher will barely review with us. Help please???
TEA [102]
So to balance an equation, you need to get the same amount of each type of element on either side of the --> . So you pretty much are given the subscripts in the equations and you need to add coefficients (just normal numbers) in front of any formula that needs it, keeping anything balance.
KCl_{3}  + O_{2}  -\ \textgreater \  KCl_{3}
turns into
2KCl_{3}+ 3O_{2} -> 2KCl_{3}

These coefficient numbers are the molar ratios, so 2 moles of KCl3 for every 3 moles of O2   so 1. 3:2

Then you can use these ratios of find out how many moles of one thing are needed if you are given the amount of another.
\frac{moles of element 1}{cofficient 1}  =  \frac{moles of element 2}{cofficient 2}
and use cross multiplication to solve for whatever you don't know


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How must atoms change so that they can join to form an ionic compound
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Atoms must gain or lose electrons in order to become ions if they are to form ionic bonds.
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Which of the following have been determined as the oldest living tree types on Earth? Select all that apply.
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Sequoia and Bristle cone pine
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A rigid tank contains 2.4 kg of helium at determine the volume
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The volume is 1.3 for
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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
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