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3 years ago

HELP PLEASE !!! Name the Salt formed When Nitric acid + Magnesium hydroxide

Chemistry

1 answer:

VladimirAG [237]3 years ago

8 0

Answer:

magnesium chloride

Explanation:

Complete step-by-step answer:

The remaining ions combine to produce the salt of neutralisation. The salt is magnesium chloride. The chemical formula for the compound is $MgC{{l}_{2}}$.

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Which is a density independent factor limiting popultion growth

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The amount the amount of space a population has to grow in would be a limiting factor.

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3 years ago

In the synthesis reaction, the white powder produced is magnesium oxide (mgo, mm = 40.3 g/mol). how many moles of magnesium oxid

gulaghasi [49]

Number of moles = mass of product / molecular mass
=mass of product (MgO) / 40.3

Since the mass of MgO is not given in the question, the correct answer choice cannot be given. However, proceeding witht eh above formula will enable you to find the correct number of moles given the mass of MgO.

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3 years ago

What is the job of these macromolecules

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Answer:

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4 years ago

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

3 0

3 years ago

How many moles are contained in a 4.67-l sample of gas at 30°c and 199 kpa?

ohaa [14]

Pv=nRT
where,p=199, R(constant)=8.314, V=4.67 T=30C=293K
n=pv/RT=0.38 moles

4 0

3 years ago

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