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Naddika [18.5K]
3 years ago
8

Wind patterns choose all that apply

Chemistry
1 answer:
kumpel [21]3 years ago
4 0
I think the answer is B.
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Sodium sulfide reacts with hydrochloric acid to produce hydrosulfuric acid and sodium chloride. (you need to write and balance t
VikaD [51]

Answer:

2.5 moles of NaCl

Explanation:

The balanced chemical reaction equation is shown in the image. Since it takes 2 moles of Hydrochloric acid to form two moles of sodium. Chloride, then 2.5 moles of hydrochloric acid should also form 2.5 moles of sodium chloride according to the balanced reaction equation.

4 0
3 years ago
Can someone please answer this?
Amanda [17]

Answer:-

0.229 L

Explanation:-

Molar mass of AgBr = 107.87 x 1 + 79.9 x 1

=187.77 grams mol-1

Mass of AgBr = 150 grams

Number of moles of AgBr = 150 grams / 187.77 gram mol-1

= 0.8 mol

The balanced chemical equation is

NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)

From the equation we can see that

1 mol of AgBr is produced from 1 mol of AgNO3.

∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.

Strength of AgNO3 = 3.5 M

Volume of AgNO3 required = Number of moles / strength

= 0.8 moles / 3.5

=0.229 L

8 0
3 years ago
13. Which statement best describes an element? *
Natasha2012 [34]

Explanation:

Distinguish chemical substances from mixtures

Key Points

Matter can be broken down into two categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components.

A chemical substance is composed of one type of atom or molecule.

A mixture is composed of different types of atoms or molecules that are not chemically bonded.

A heterogeneous mixture is a mixture of two or more chemical substances where the various components can be visually distinguished.

A homogeneous mixture is a type of mixture in which the composition is uniform and every part of the solution has the same properties.

Various separation techniques exist in order to separate matter, including include distillation, filtration, evaporation and chromatography. Matter can be in the same phase or in two different phases for this separation to take place.

Terms

substanceA form of matter that has constant chemical composition and characteristic properties. It is composed of one type of atom or molecule.

elementA chemical substance that is made up of a particular kind of atom and cannot be broken down or transformed by a chemical reaction.

mixtureSomething that consists of diverse, non-bonded elements or molecules.

4 0
3 years ago
Helen recorded the following data about the half-life of a radioisotope. Radioactive Decay of Radioisotope A Grams of Radioactiv
MissTica

Answer:

Line graph

Explanation:

I did it on Study Island

7 0
3 years ago
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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