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ohaa [14]
3 years ago
9

PLEASE PLEASE HELP

Chemistry
1 answer:
Eva8 [605]3 years ago
8 0
Try googleing it that is how i got some of mine

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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
I need the correct answer
zalisa [80]

Answer:

d.

Explanation:

They can be spread through contaminated surfaces.

6 0
3 years ago
Read 2 more answers
What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 15.0% NaOH by mass?
GalinKa [24]
Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH

Now, we apply the definition of molality to calculate the molality of the solution:

mNaOH= \frac{moles NaOH}{kg_{solvent}}=  \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m

So, the molality of this solution is 4,41 m

Have a nice day!
4 0
3 years ago
Which crystalline solds generally have the highest melting and boiling points?
strojnjashka [21]
Metallic (w) with a boiling point of 566”
6 0
3 years ago
1. In this experiment, what property of NaCl is used to separate it from the other two components? Is this a chemical or physica
irinina [24]

Answer 1) In the mixture of sand, NaCl and CaCO_{3} first separation was done by dissolving the mixture in water. In the first step the NaCl will get dissolved whereas, CaCO_{3} will be undissolved and sand will settle at the bottom. Here, the physical property of solubility of NaCl in water is taken into consideration. The collected water is then filtered off and evaporated to get the NaCl back from the mixture with some loss. No chemical change occurs in case of NaCl extraction from water.


Answer 2) When CaCO_{3} was to be removed from the undissolved part. The chemical property of CaCO_{3} was used. Where it gets dissolved in acidic medium. And this way we can extract CaCO_{3} and remove sand from it. We change the chemical composition of CaCO_{3} by adding HCl to the mixture and dissolve CaCO_{3} to form CaCl_{2} which gets dissolved into the solution of HCl. Here, first decantation occurs and then extraction is done. Second, extraction is done using potassium carbonate in it which separates CaCl_{2} from sand.


Answer 3) Here, Unknown mixture of salt and sand weighed =7.52 g (before washing);


Unknown mixture of salt and sand weighed =3.45 g (after washing);


To calculate the amount of salt in it, we can simply subtract the values of before and after washing change in weights.


The property of NaCl being soluble in water it will go away with washing leaving behind the sand only, after washing.


So, Weight of salt was = 7.52g - 3.45 g = 4.07g


To find the percentage of sand that was mixed with salt =

(3.45 g sand / 7.52g of mixture) X 100 = 45.9% sand was mixed with salt.


To verify whether the correct percentage of salt and sand was calculated we can recheck the value for salt as well.

(4.07 g of salt / 7.52g of mixture) X 100 = 54.1% salt


On adding we get, 45.9% + 54.1% = 100%.


Which confirms that the calculations are correct.

4 0
3 years ago
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