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3 years ago

5.22x10-3 = [? ]x 10-2

Chemistry

1 answer:

levacccp [35]3 years ago

4 0

Answer:

5.12

Explanation:

5.22 x 10 - 3 = [?] x 10 - 2 (move - 2 to the other side with changed sign)

5.22 x 10 - 3 + 2 = [?] x 10 (divide both sides by 10)

[?] = (5.22 x 10 - 1) / 10

[?] = 5.12

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liubo4ka [24]

Answer:

B. Bottom to top

Explanation:

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2 years ago

The solubility of k2cr2o7 in water is 125 g/l at 20 °c. a solution is prepared at 20 °c that contains 6.0 grams of k2cr2o7 in 50

Pachacha [2.7K]

The solubility is the guide to the maximum amount of solute that can be dissolved in a certain amount of solvent at a certain temperature to make a saturated solution. Any amount less than this would result to unsaturated, while any amount more would result to saturated.

6 g/(50 mL * 1 L/1000 mL) = 120 g/L

Since it is less than the solubility of 125 g/L, then <em>this solution is unsaturated</em>.

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3 years ago

Density is the ratio of mass to volume. Another way to say this is

Oliga [24]

Answer:

my teacher would let u say the density to mass is the ratio but i think everyone's different

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3 years ago

A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh

Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0

3 years ago

Why must the ph ammonia buffer solution be stored and dispensed in the fume exhaust hood?

Vadim26 [7]

On the other hand ammonia is a very dangerous chemical which has a pungent smell and effect the eyes of the user. Thus it kept always in the fume exhaust hood for storing and dispensing function.

The pH of ammonia buffer contains ammonium hydroxide (NH₄OH) and a salt of ammonia with a strong acid like (HCl) which produces, ammonium chloride (NH₄Cl) mixture. The evaporation rate of ammonia is so high at room temperature thus on opening of the buffer solution the ammonia get evaporated very fast and the concentration of ammonia decreases which affect the pH of the buffer solution.

Thus the reason to put ammonia buffer in fume hood is explained.

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3 years ago