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Fofino [41]
3 years ago
8

Gasoline has a density of .74g/mL. How many milliliter if gasoline have a mass of 2.5kg?

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0

Answer: 3378.4mL

Explanation:

density of Gasoline = 0.74g/mL.

Volume of gasoline in milliliter = ?

mass of gasoline = 2.5kg

(Since density is in grams per millilitre, convert mass in kilograms to grams)

If 1kg = 1000grams

2.5kg = 2.5 x 1000 = 2500grams

Recall that density is obtained by dividing the mass of a substance by its volume

i.e Density = Mass / Volume

0.74g/mL = 2500grams/Volume

Volume = (2500 grams / 0.74g/mL)

Volume = 3378.4mL

Thus, the volume of gasoline is 3378.4mL

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The km of an enzyme is 5. 0 mm. Calculate the substrate concentration when this enzyme operates at one‑quarter of its maximum ra
Makovka662 [10]

The substrate concentration of the enzyme operating at one‑quarter of its maximum rate is = 0.333.

Relationship between Km and substrate concentration is -

Km is the concentration of substrate.It allows the enzyme to achieve half Vmax. High Km enzyme requires a higher concentration of substrate to get Vmax. Since, Km is a constant. If the substrate concentration is increased, it has no effect on it.

An enzyme with a high Km has a low affinity for its substrate. The substrate concentration Km corresponds to the substrate concentration.

The substrate concentration at which the reaction rate of the enzyme-catalyzed reaction is half of the maximum reaction rate Vmax.

The equation is:

                <em>V₀ = </em><u><em>Vmax [S] </em></u>

<em>                          [S] + Km</em>

Here,

V₀ is initial rate,

Km is the dissociation constant between the substrate and the enzyme,

Vmax is the maximum rate, and

S is the concentration of substrate.

taking fraction of V₀ and Vmax :

<u><em>    V₀    </em></u><em> = </em><u><em>     [S</em></u><em>]</em><u><em>     </em></u><em>  </em>

<em>Vmax      [S] + Km</em>

<u><em>   </em></u><u>V₀   </u> =   <u>   0.5Km   </u>  = 0.333

Vmax       1.50 + Km

Therefore, the substrate concentration of this enzyme operating at one‑quarter of its maximum rate is = 0.333.

To learn more about substrate concentration,

brainly.com/question/18237939

#SPJ4

3 0
1 year ago
A 1.897g sample of Mg(HCO3)2 was heated and decomposed. When the sample
svlad2 [7]

1.5 % is the percent yield in the reaction.

Explanation:

Given that:

original mass of the sample used in reaction = 1.897 grams

product formed after decomposition = 1.071 grams

The reaction for the decomposition:

Mg(HCO3)2 (s) ⇒ CO2 (g) + H2O (g) + MgCO2 (s)

It says that 1 mole of Mg(HCO3)2  yielded 1 mole of  MgCO2  on decomposition

68.31 grams/mole or 68.31 grams of MgCO2 is formed

percent yield = \frac{actual yield}{theoretical yield} x 100

putting the values in the equation:

percent yield = \frac{1.071}{68.31}

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   PERCENT YIELD = 1.5 %

8 0
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