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wel
2 years ago
15

Jane set up an experiment with 7 types of dog food and 15 dogs. Each day she fed a different brand of dog food to the 15 dogs. H

er
results are recorded in the bar graph above.
What might be the question Jane is asking in her experiment?
A)
What dog food cost the most?
B)
Which dog food do dogs like the best?
Which dog food has the most real meat?
D)
How many dogs only like to eat in the morning?
Chemistry
1 answer:
bixtya [17]2 years ago
4 0

Answer:

I believe it's either D or A I did it in my math

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If a 130-kg car is traveling at a velocity of 23 m/s, what is the kinetic energy of the car?​
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KE = 1/2 * m * v^2
KE = 1/2 * 130 * 23^2
KE = 34385J
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3 years ago
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GIVING BRAINLIEST!! PLEASE HELP
PIT_PIT [208]

Answer:

d

Explanation:

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2 years ago
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Cedrick & Astrid titrated a 10.00 mL aliquot of grapefruit juice with a 0.171 M NaOH solution to the end point. They calcula
Strike441 [17]

Answer:

17 mg/ml.

Explanation:

Since NaOH was titrated against Citric acid in the grape fruit, therefore

Converting g to mg,

0.1698 g * 1000 mg/1 g

= 169.8 mg

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Mass of Citric acid in ml of juice = 169.8/10

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5 0
3 years ago
A solution contains one or more of the following ions: Hg2+2, Ba2+, and Fe2+. When potassium chloride is added to the solution,
Aloiza [94]

Answer:

Explanation:

Solubility is a measure of the ability of a given substance to dissolve in a liquid, that is, it describes the amount of solute that can be dissolved in a specific amount of solvent.  This occurs by solvation. Solvation is the process that involves the association of solvent molecules with molecules or ions of a solute. When the ions of a solvent dissolve, they are separated and surrounded by the molecules that form the solvent. The larger the size of the ion, the greater the number of molecules capable of surrounding it, so it is said that the ion is mostly solvated.

For a substance to dissolve in another the polarity of the molecules must be similar. For example, water is a polar compound and easily dissolves polar substances such as acids, hydroxides and inorganic salts and polar organic compounds. Although there are also exceptions. There are highly polar inorganic compounds that are insoluble in water, such as carbonates or sulphides.

Taking into account the aforementioned, the solubility rules were stated. One of these rules mentions that all chlorides (Cl-) are soluble except those of Hg₂²⁺, among others.  This means that by adding KCl to the mixture, the compound Hg₂Cl precipitates through the net ionic equation:

Hg₂²⁺ (aq) + Cl⁻ (aq) ⇒ Hg₂Cl (s)

Then the solution is filtered with the precipitate so that no more Hg₂²⁺ is present in the solution.

Another rule of solubility indicates that sulfates (SO₄²⁻) are all soluble except those of Ba²⁺, among others.  That is, between the sulfate and Ba²⁺ precipitate should form. But since it is indicated that potassium sulfate is added to the remaining solution without producing precipitate, it is possible to say that the sulfate is actually reacting with Fe²⁺. This, due to the solubility rule, is soluble with sulfate, producing no precipitate. So, since there is no precipitate, there is no net ionic equation. And there should be no Fe + 2 in the solution due to its reaction with sulfate.

Finally, another solubility rule indicates that all carbonates (CO₃²⁻), phosphates (PO₄³⁻), arsenates (As⁺⁵) and chromates (Cr₂O₄²⁻) are insoluble, except those of Group IA and those of NH₄⁺. Then  when potassium carbonate is added to the remaining solution, a precipitate is formed, product of the net ionic equation between Ba²⁺ and carbonate:

Ba²⁺ (aq) + CO₃²⁻ (aq) ⇒ BaCO₃ (s)

7 0
2 years ago
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A 1.78−L sample of hydrogen chloride (HCl) gas at 2.09 atm and 22°C is completely dissolved in 699 mL of water to form hydrochlo
patriot [66]

Answer:

M=0.960M

Explanation:

Hello,

In this case, it is firstly necessary to compute the dissolved moles of hydrogen chloride into the water as shown below:

n_{HCl}=\frac{PV}{RT}=\frac{2.09atm*1.78L}{0.082\frac{atm*L}{mol*K}*295.15K}=0.671molHCl

Thus, the molarity is computed as shown below:

M=\frac{0.671mol}{0.699L}=0.960M

Wherein no change in volume is considered, therefore the volume of the solution was the same volume of water.

Best regards.

5 0
3 years ago
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