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3 years ago

What are units used for specific gravity? its 20 points ;)

Chemistry

1 answer:

Amanda [17]3 years ago

6 0

Answer:

Explanation: Specific gravity is a dimensionless quantity; that is, it is not expressed in units. To find the sp gr of a solid or liquid, you must know its density in kilograms per meter cubed (kg/m3) or in grams per centimeter cubed (g/cm3). Then, divide this density by the density of pure water in the same units.

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What two factors most affect the pitch of the sounds made by a person's vocal cords? A. the volume and health of the person's lu

dusya [7]

Answer:

The pitch of the sound a person makes is determined by several factors, including the size and tension of his or her vocal cords.

Explanation:

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3 years ago

Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50∘C

trapecia [35]

Answer:(16)(2000)m/s=32000m/s

Explanation: A diatomic oxygen has a molar mass of 16

8 0

3 years ago

Density is ___ to pressure

Ann [662]

Proportional to pressure

5 0

3 years ago

Some friends are trying to make wine in their basement. They've added yeast to a sweet grape juice mixture and have allowed the

Mekhanik [1.2K]

Answer:

The yeast respired aerobically

Explanation:

The expectation here was that the yeast was going to put the sugar in the grape juice through fermentation, which is an anaerobic process that results in alcohol as one of the products. However since no alcohol was found in this particular example, it is fair to assume that maybe there was oxygen in teh muxture and therefore the yeast respired aerobically, producing water and carbon dioxide as products.

8 0

3 years ago

The enzyme, phosphoglucomutase, catalyzes the interconversion

Fittoniya [83]

Answer:

$K_{eq$ = 19

ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J

ΔG° of the reaction under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate ⇄ Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant $K_{eq$ can be calculated as:

$K_{eq$ $= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}$

$K_{eq$ $= \frac{0.95}{0.05}$

$K_{eq$ = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant $K_{eq$ can be calculated as:

$K_{eq$ $= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}$

$K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}$

$K_{eq} =$ 0.01279816514 M

$K_{eq} =$ 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0

3 years ago