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3 years ago

How do I balance a chemical equation?

Chemistry

1 answer:

bogdanovich [222]3 years ago

8 0

Answer:

Concept: Chemical Analysis

Start by taking inventory of the elements that you have
Make a list, one for the right side and another for the left side
Then add coefficients to the elements to the right or left side to balance out the equation

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Write application of isotopes .

SCORPION-xisa [38]

Answer:

In the chemical industry.

6 0

2 years ago

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How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----> K3PO4 + 3

8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

$\frac{3}{1} =\frac{y}{0.02256} \\$

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

5 0

3 years ago

The half life for the decay of radium is 1620 years what is the rate constant

Lilit [14]

Decay is a type of degradation reaction and thus is considered a first order reaction. thus the formula goes like this.

rate constant= 0.693/half life

so here...

rate constant= 0.693/1620 year^-1

3 0

3 years ago

A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

NeTakaya

Answer:

solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 $^{0}\textrm{C}$ can be calculated using the information given.

Let's assume solubility of X in water at 17.0 $^{0}\textrm{C}$ is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

or, y = $\frac{3.96}{36}$ = 0.11

Hence solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

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3 years ago

Complete this sentence. If mass remains the same while the volume of a substance ________, the density of the substance will____

Andre45 [30]

<span> If mass remains the same while the volume of a substance increases, the density of the substance will decrease. The increase in the volume is called expansion, while the decrease in the volume is called contraction.
</span><span>The density measures how much matter (mass) there is in a given amount of space (volume). So, if the mass stays the same and the volume increases then there is the same amount of mass ad before but in bigger volume so the density is decreased. </span>

4 0

3 years ago

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