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3 years ago

Magnesium sulfide + nitric acid net ionic

Chemistry

1 answer:

Oxana [17]3 years ago

8 0

Answer:

MgSO_3(s) + 2 HNO_3(aq) rightarrow Mg^2+(aq) + 2 NO_3^-(aq) + SO_2(g) + H_2O(l) b.

Explanation:

The balanced net ionic equation for the reaction of magnesium sulfite with nitric acid is a. MgSO_3(s) + 2 HNO_3(aq) rightarrow Mg^2+(aq) + 2 NO_3^-(aq) + SO_2(g) + H_2O(l) b.

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eimsori [14]

As volume increases, the temperature will increase.

As volume decreases, the temperature will decrease.

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3 years ago

What type of product is formed when acids are added to some ionic compounds

mars1129 [50]

Answer:

salt

Explanation:

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3 years ago

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

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R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

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Therefore, the mass of $NaN_3$ required is, 166.4 grams.

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3 years ago

Using your value of Ksp, and starting with an equilibrium system consisting of a saturated solution of calcium hydroxide, predic

Ugo [173]

Answer:

(A) $HNO_{3}$ Solubility will increase.

(B) $NaOH$ Solubility will decrease.

(C) $Ca(NO_{3} )_{2}$ Solubility will decrease.

Explanation:

a) According to Le Chatelier's Principle when a change is introduced in a reaction system at equilibrium, the system responds by the reaction shifting in the direction to counter the change introduced.

When the change introduced is addition of acid, some $OH^{-}$ are consumed due to it and the system would respond by reaction shifting towards creation of more $OH^{-}$ in the system by increased dissolution. Thus solubility increase.

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(c) The change introduced is addition of $Ca^{2+}$ which will be countered by reaction shifting to side that decreases increased $Ca ^{2+}$ , which happens by some $Ca(OH)_{2}$ precipitating back. Hence solubility decreases.

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4 years ago

How many moles of carbon dioxide are present in 14g of co2

Tomtit [17]

0.32 moles

You just do 14/ 44.01 (the atomic mass of carbon + oxygen's atomic massX2)

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3 years ago