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Kaylis [27]
3 years ago
5

Calculate the density of an object that has a mass of 43 g and a volume of 56.0 ml.

Chemistry
1 answer:
Yanka [14]3 years ago
5 0
I will try to do it but not more than 70.4

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Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5
FinnZ [79.3K]
The answer for this issue is: 
The chemical equation is: HBz + H2O <- - > H3O+ + Bz- 
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz] 
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x. 
Accept that x is little contrasted with 0.5 M. At that point, 
Ka = 6.4X10^-5 = x^2/0.5 
x = [H3O+] = 5.6X10^-3 M 
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make) 
3 0
4 years ago
Write balanced, net ionic equations for the following precipitation, acidn base, or gasn forming reactions. Include states of ma
dybincka [34]

Answer:

Explanation:

a ) 2FeCl₃ + 3Li₂S  = Fe₂S₃ ( s )  + 6 LiCl

2Fe⁺³ + 6Li ⁻ + 6Cl⁻ + 3S⁻² = 6Li + 6Cl⁻ + Fe₂S₃ ( s )

b  )

3CH₃COONa +( NH₄)₃PO₄ = 3CH₃COONH₄ + Na₃PO₄

3CH₃COO + 3Na⁺ +  3NH₄⁻  +  PO₄⁺³  = 3CH₃COO⁻ +3NH₄⁺ + Na₃PO₄

c )

HClO₄  + KOH = kClO₄ + H₂O

H ⁺ + ClO₄⁻ +  K⁺ + OH⁻  =  k⁺  ClO₄⁻  + H₂O

d )

NH₄OH + HNO₃ = NH₄NO₃  + H₂O

NH₄⁺ + OH⁻ + H⁺ + NO₃⁻ = NH₄⁺ + NO₃⁻ +  H₂O

e )

HNO₂ + KOH = KNO₂ + H₂O

H⁺ + NO₂⁻ + K⁺ + OH⁻ = K⁺ + NO₂⁻ + H₂O

f ) HIO₃ + CaCO₃ ( s ) = Ca( IO₃ )₂ + H₂CO₃

H⁺ + IO₃⁻ + CaCO₃ ( s ) = Ca( IO₃ )₂ + H₂CO₃

g )

c ) is strong acid and strong base

d ) is weak base and strong acid

e ) weak acid and strong base

f ) Strong acid and basic salt

5 0
3 years ago
Which statement describes the water gas particles in the air above the cup compared with the water liquid
Law Incorporation [45]

Answer:

The particles move faster and are far apart

Explanation:

A substance may exist in three states of matter; solid, liquid and gas.

In the solid state, there is very strong intermolecular forces between the particles of the substance. They can only vibrate or rotate about their mean positions but can not translate.

In the liquid state, the particles of the substance have a greater degree of freedom than in the solid. The magnitude of intermolecular forces is lower than in solids, the molecules can move at low speeds.

In a gas, the molecules are separated from each other with negligible intermolecular interaction hence they move at very high speed.

Therefore, for the water gas particles in the air above the cup; the particles move faster and are far apart.

8 0
3 years ago
What quantities are conserved when balancing a chemical reaction?
EleoNora [17]
The mass in a chemical reaction remains (mostly) the same.

(except for radiation/nuclear fission, in which mass gets converted into energy)
8 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
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