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2 years ago

2. A chemical analysis of a sample provides the following elemental data:

Chemistry

1 answer:

Vadim26 [7]2 years ago

3 0

Answer:

C3 H6 O2

Explanation:

first divide their mass by their respective molar mass, we get:

30.4 moles of C

61.2 moles of H

20.25 moles of O

now divide everyone by the smallest one of them then we get

C= 1.5

H= 3

O= 1

since our answer of C is not near to any whole number so we will multiply all of them by 2

so,

C3 H6 O2 is our answer

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What is the density of an object if its mass is 10 g and a volume of 2 mL?

Lilit [14]

Hi there! Let's solve this problem shall we!

⠀Volume = 10g

Mass = 2 mL

In this specific problem, they are asking us to find the density of the object. So, using the information given to us (volume and mass), let's solve the problem!

Now, if you remember, D = M ÷ V

So, let's fill in the blanks!

D = Our unknown value

M = 2mL

V = 10g

Here is the filled out formula:

D = M ÷ V

D = 2mL ÷ 10g

D = 5 g/mL

*Make sure you put the units for your final solution!*

4 0

2 years ago

The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.

maw [93]

CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L

7 0

3 years ago

When did oxygen first appear in earth's atmosphere??

ad-work [718]

the atmosphere holds about 21 per cent oxygen. Over the Earth’s 4.6 billion year history, oxygen did not appear in the atmosphere until perhaps about 2.5 billion years ago. Since then, oxygen levels have fluctuated in tandem with global geological and biological events, such as mass extinctions.

7 0

3 years ago

Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca

Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

$k=Aexp(-Ea/RT)\\$