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inna [77]
3 years ago
7

What is the density of a block of marble that occupies 277 cm3 and has a mass of 928 g? Answer in units of g/cm3 .

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
8 0
V=277cm^{3}\\
m=928g\\\\
d=\frac{m}{V}=\frac{928g}{277cm^{3}}\approx3,35\frac{g}{cm^{3}}
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Define the following symbols that are encountered in rate equations: [A]0, t1/2 [A]t, k.
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[A]0= Initial concentration
t1/2= half life
[A]= final concentration
k= rate constant
5 0
3 years ago
If aluminum has a density of 2.7 g/cm³, what is the volume of 2.7 grams of aluminum?​
Lorico [155]

Answer:

Solution Density of aluminium = 2.7 g/Cm 3 In kg/ m 3 = 27 × 1000 10 =2700 kg/ m 3

Explanation:

Not much of one

4 0
1 year ago
In matter results in a change in its identity and properties.
FrozenT [24]

Answer:

The chemical change in matter is results in a change in its identity and properties.

Explanation:

There are toe type of changes physical change and chemical change.

The chemical change change in matter lead to lost its identity and properties.

Chemical change:

The changes, that occur due to change in the composition of a substance and result in a different compound is known as chemical change.

These changes are irreversible

These changes occur due to chemical reactions

These may not be observed with naked eye

Example:

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Boiling of egg: that change the chemical composition of protein in the egg .

The reaction of Hydrogen and oxygen:  

H 2 (g)  + O 2  (g) -------------------------------------> 2H 2O  (l)

7 0
3 years ago
What is the quantity of atoms in KMnO4 ?
tankabanditka [31]

Answer:

Potassium permanganate has a molar mass of 158.04 g/mol. This figure is obtained by adding the individual molar masses of <em><u>four oxygen atoms</u></em>, <em><u>one manganese atom</u></em> and <em><u>one potassium atom</u></em>

Explanation:

6 0
2 years ago
Read 2 more answers
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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