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4 years ago

What coefficient is assumed if no coefficient written before a formula in a chemical equation

Chemistry

1 answer:

Mice21 [21]4 years ago

7 0

Answer:

Explanation:

Consider the following reaction:

Phosphorus burn in the presence of air and produced diphosphorus pentoxide.

Chemical equation:

P₄ + O₂ → P₂O₅

Balanced chemical equation:

P₄ + 5O₂ → 2P₂O₅

Equation is balanced because there are four phosphorus atoms ans ten oxygen atoms in both side of equation.

Coefficient with reactant and product:

P₄ 1

O₂ 5

P₂O₅ 2

There is no coefficient written before P₄ it means only one P₄ is present.

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What coefficient for O2 demonstrates the law of conservation of mass?

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Explanation:

The coefficient for O is 2 and this is an example of a combustion reaction. With the help of the coefficient 2 infront of oxygen, this equation now demonstrates law of conservation of mass.

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3 years ago

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3 years ago

Solid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest

oksian1 [2.3K]

Answer

Hello

I think the reaction is like this FeO+OFe²O³

And the balance reaction is 2Fe+OFe²O³

Explanation:

At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.

Finally our reaction balanced.

Good luck

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3 years ago

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha

lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

$\frac{60}{100}\times 3.6 = 2.16 L$

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator =50% of 3.6 L = $\frac{50}{100}\times 3.6=1.8 L$

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired volume

2.16 L - 60% of( x) = 1.8 L

$2.16 L-0.6x=1.8 L$

$x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L$

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

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4 years ago