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bagirrra123 [75]
3 years ago
9

The weight of an object never changes. True False

Chemistry
2 answers:
goblinko [34]3 years ago
8 0

Answer: False depending on what you do or what happens to the object but true if nothing happens to it

Explanation:

sertanlavr [38]3 years ago
6 0

the answer to your question is

true

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A story on chemical reaction or electric current
Anuta_ua [19.1K]

Answer:

a process that involves rearrangement of the molecular or ionic structure of a substance, as opposed to a change in physical form or a nuclear reaction

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3 years ago
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The modern periodic table is organized by increasing ______________ ___________.
professor190 [17]

Answer:

atomic number.

Explanation:

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2 years ago
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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
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4 0
3 years ago
Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

V = 38.75 L

3 0
3 years ago
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