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Drupady [299]
2 years ago
13

Pls help!!!!!! In your own words, explain how chemists properly represent the law of conservation of matter in their chemical

Chemistry
1 answer:
WINSTONCH [101]2 years ago
5 0

Answer:

They represent it by ensuring that the number of atoms of each element (matter) in the reactant side is the same as the product side

Explanation:

The law of conservation of matter stated that matter can neither be created nor destroyed. Chemical equations involve combining atoms of elements. The compounds combined by chemists are called REACTANTS while the produced compounds are called PRODUCTS.

In order to conform to the law of conservation of matter, the same quantity of matter present in the reactants must be present in the products. This means that the number of atoms of each element (matter) in the reactant side must be the same as the product side. For example;

C6H12O6 + 6O2 → 6CO2 + 6H2O

In this chemical equation for photosynthesis, number of atoms in the reactant side (6 carbon, 12 hydrogen, 18 oxygen) are the same as that in the product side (6 carbon, 12 hydrogen, 18 oxygen), hence, this obeys the law of conservation of mass.

In a nutshell, chemists chemists properly represent the law of conservation of matter in their chemical equations by making sure that same number of atoms of reactants is present in the products.

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Metals are found on the left-hand side of the periodic table. True or False
Tanya [424]

Answer:

true pls mark brainiest

Explanation:

5 0
3 years ago
What is the effect of adding heat to a gas at constant pressure?
Anna007 [38]
The pressure will continue to build up eventually causing a release of pressure or an explosion.
8 0
2 years ago
Amphetamine (C9H13N) is a weak base with a pKb of 4.2.
QveST [7]

Weak base: [OH⁻] = √Kb.C

pKb = 4.2

\tt Kb=10^{-4.2}

c = concentration

MM Amphetamine (C9H13N) = 135.21 g/mol

c = 215 mg/L = (0.215 g : 135,21 g/mol) / L = 0.00159 mol/L = 1.59  x 10⁻³ mol/L

\tt [OH^-]=\sqrt{10^{-4.2}\times 1.59\times 10^{-3}}=3.17\times 10^{-4}

pOH = 4 - log 3.17

pH = 14 - (4 - log 3.17)

pH = 10 + log 3.17 = 10.50

4 0
2 years ago
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
2 years ago
Some gas was collected when the pressure was 760mmHg. It occupied 500mL when the pressure was increased to 800mmHg. What was the
Ivan

Answer:

<h3>The answer is 526.32 mL</h3>

Explanation:

To find the original volume we use the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the original volume

V_1 =  \frac{P_2V_2}{P_1}  \\

From the question we have

V_1 =  \frac{500 \times 800}{760}  =  \frac{400000}{760}  =  \frac{40000}{76}  \\  = 526.315789...

We have the final answer as

<h3>526.32 mL</h3>

Hope this helps you

6 0
2 years ago
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