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2 years ago

Which of the following is a valid conversion factor?

Chemistry

1 answer:

Novosadov [1.4K]2 years ago

7 0

Answer:

100cg-1g

Explanation:

This is because for:

10m=100dm

1ml = 0.001L

1000000cm3=1L

100cg=1g

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A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

likoan [24]

Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

To Find :

The volume of the gas after it is heated.

Solution :

Since, their is no information about pressure in the question statement let us assume that pressure is constant.

Now, we know by ideal gas equation at constant pressure :

$\dfrac{V_1}{V_2} = \dfrac{T_1}{T_2}\\\\\dfrac{3.82}{V_2}= \dfrac{204.9}{304.8}\\\\V_2 = \dfrac{304.8}{204.9} \times 3.82\\\\V_2 = 5.68 \ L$

Hence, this is the required solution.

3 0

3 years ago

What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +

nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ + S₂O₃²⁻ + H₂O → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻ Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0

3 years ago

The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The

dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

$_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e$

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

$_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e$

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0

3 years ago

1.A telescope is a tool that helps people observe distant objects. In the simplest telescope design, incoming light hits a lens

Dahasolnce [82]

Answer:

it is refraction

Explanation:

7 0

3 years ago

Which of these 3 molecules is most polar? And which is the least polar? H2O, H2S, or SO2? Explain.

stepan [7]

Answer:so2

Explanation:

5 0

3 years ago