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Marrrta [24]
3 years ago
11

Someone please help me and answer my question asap. i need to turn my assignment in soon... please.

Chemistry
1 answer:
Ksenya-84 [330]3 years ago
8 0

Answer:

Explanation:

a. the salt produced would be Mg3N2(magnesium nitride)

b. magnesium loses 2 electron to form Mg2+ ion and nitrogen gains 3 electron to form n3-

when several of these ions come together 3 Mg2+ ion combine with 2 n3- ion to form Mg3N2 thus Mg getting six electron from nitrogen to form a ionic bond.

c. the reaction is not balanced Mg + N2 = Mg3n2

to make it balanced the reaction should be 3 Mg + N2 = Mg3N2.

the reaction was not balanced before because the number of Mg on both side of the reaction was not equal.

d. magnesium nitrate has formula Mg(NO3)2 is formed when Mg combines with nitrogen and oxygen Mg + N2 + o2

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Alika [10]

Answer:

it would break when it hits the ground

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A mixture of helium, nitrogen and oxygen has a total pressure of 821 mmHg. The partial pressure of helium is 105 mmHg, and the p
skad [1K]

Answer:

Total partial pressure, Pt = 821 mm Hg

Partial pressure of Helium, P1 = 105 mm Hg

Partial pressure of Nitrogen, P2 = 312 mm Hg

Partial pressure of Oxygen, P3 = ? mm Hg

According to Dalton's law of Partial pressures,

Pt = P1 + P2 + P3

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4 0
2 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
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Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
Which of these equations is balanced? a. H2SO4 + 2Al → Al2(SO4)3 + H2 b. 2KCl + Pb(NO3)2 → 2KNO3 + PbCl2
Lapatulllka [165]
The answer is B, you just check if it is the same on the left and right side

A:
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B is therefore correct as both sides add up
6 0
3 years ago
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