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3 years ago

Write a balanced chemical equation for the standard formation reaction of solid sodium orthosilicate

Chemistry

2 answers:

Mumz [18]3 years ago

8 0

Equation for the standard formation of solid sodium orthosilicate:
4 Na(s) + Si(s) + 2 O₂(g) → Na₄SiO₄(s)

Brums [2.3K]3 years ago

3 0

Answer : $Si_{(s)} + 4NaOH_{(aq)}---> Na_{4}SiO_{4}_{(s)} + 2H_{2}_{(g)}$

Explanation :

Silicon in presence of aqueous sodium hydroxide undergoes single displacement/replacement reaction and forms solid sodium orthosilicate which is a stable compound.

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David's body breaks down the sandwich into small enough molecules that can be absorbed through his intestinal lining and then in

viva [34]

break down food into smaller molecules the body can use; absorb molecules into blood and carry them throughout the body; eliminate wastes from the body

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3 years ago

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A ninja motorbike travelling at 55 m/s takes 10 sec to come to rest . What is its deceleration rate ?

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Answer:

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2 years ago

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A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what

bulgar [2K]

Answer:

The mass of KF in the mixture is 2.77 gms.

Explanation:

Given;

Total weight of mixture (LiF+KF)=5.32gms

Let, mass of KF in the mixture = x gms

⟹ mass of LiF in mixture =(5.32-x)gms.

We know that :

Atomic weight of F=19gms.

Atomic weight of Li =7gms.

Atomic weight of K = 39 gms.

moles=mass/(molecular weight)

Thus, moles of KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

Thus,

moles of F in KF=moles of KF=x/58 ---(1)

moles of F in LiF =moles of LiF= (5.32-x)/26---(2)

From (1) & (2),

Total moles of Fluorine

=(x/58)+((5.32-x)/26)

Hence,

total weight of Fluorine in sample = moles*Atomic weight

=((x/58)+((5.32-x)/26))*19gms.

=3.12 gms.---(given)

Now, solving the equation for x,

26x +(5.32*58)-58x

=3.12*58*26/19

22x=308.56-247.62

x=60.94/22

=2.77 gms. (Answer)

Thus, the mass of KF in the mixture is 2.77 gms.

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3 years ago

How many moles are in 155 grams of aluminum

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Answer:

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Explanation:

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3 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

$24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of 0.541284 moles of MgO:

$m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}$ .

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2 years ago