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omeli [17]
3 years ago
6

Naomi has increased the pressure on a solution of liquid and gas in a closed container. What will this do to the gas in her solu

tion?
A. increase the amount

B. decrease the amount

C. increase the temperature

D. decrease the temperature
Chemistry
2 answers:
dsp733 years ago
6 0

Answer:

A.

Explanation:

dedylja [7]3 years ago
5 0

Answer:

increase the amount.

Hope this helps =)

Explanation:

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What does the oxidizing agent do in a redox reaction apex?
densk [106]
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)


6 0
3 years ago
Read 2 more answers
Hey! I was wondering how to get the solution to this question!
zzz [600]

Answer:

6.22 × 10⁻⁵

Explanation:

Step 1: Write the dissociation reaction

HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺

Step 2: Calculate the concentration of H⁺

The pH of the solution is 2.78.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M

Step 3: Calculate the molar concentration of the benzoic acid

We will use the following expression.

Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution

Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M

Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid

We will use the following expression.

Ka = [H⁺]²/Ca

Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵

3 0
3 years ago
What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3
Dmitriy789 [7]
d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\
V=4,2cm^{3}\\\\
m=dV=7,784\frac{g}{cm^{3}}*4,2cm^{3}\approx 32,7g
6 0
3 years ago
A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt
Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0
3 years ago
How do valence electrons interact between non polar covalent, polar covalent, and ionic bonds?​
iVinArrow [24]

Answer:

In non-polar covalent bonds, the electrons are equally shared between the two atoms. For atoms with differing electronegativity, the bond will be a polar covalent interaction, where the electrons will not be shared equally.

Explanation:

i did some reasherch so there^^

5 0
3 years ago
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