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3 years ago

PLEASE HELP!!!

Chemistry

1 answer:

Inga [223]3 years ago

3 0

Answer:

$d=1.5\ g/cm^3$

Explanation:

Given that,

The mass of the rock, m = 15 g

The volume of the rock, V = 10 cm³

We need to find the density of the rock. The density of an object is equal to the mass per unit volume such that,

$d=\dfrac{m}{V}\\\\d=\dfrac{15\ g}{10\ cm^3}\\\\d=1.5\ g/cm^3$

So, the density of the rock is equal to $1.5\ g/cm^3$ .

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prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

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[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

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Therefore :

Kcat = Vmax/ [E]t

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Hope that helps!

4 0

3 years ago

Round off 4.5778 x 10mg to three significant figures.

lakkis [162]

Answer:

Explanation:

= 45.8

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3 years ago

A mixture of 5L of H2 and 3L O2 reacts to form H2O (g) at constant T and P .Find the

slamgirl [31]

The volume of H₂O = 5 L

<h3>Further explanation</h3>

Given

5L of H₂ and 3L O₂

Reaction

2H₂ (g) + O₂(g) ⇒2H₂O(g)

Required

The volume of H₂O

Solution

Avogadro's hypothesis:

In the same T,P and V, the gas contains the same number of molecules

So the ratio of gas volume will be equal to the ratio of gas moles

mol H₂ = 5, mol O₂ = 3

Find limiting reactants

From equation, mol ratio H₂ : O₂ = 2 : 1, so :

$\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)$

Find volume H₂O

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

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3 years ago

Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w

jek_recluse [69]

The question is incomplete, here is the complete question:

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$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

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