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3 years ago

Write the chemical equation for the reaction in which silver carbonate is formed

Chemistry

1 answer:

OverLord2011 [107]3 years ago

8 0

Explanation:

Na2CO3+2AgNO3------>Ag2CO3+2NaNO3

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Which water carries the most oxygen in streams and rivers?

lbvjy [14]

the answer is fresh water

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4 years ago

Crude oil may contain hundreds of different types of hydrocarbons. Some examples include: Butane(C4H10) Dodecane (C12H26) Octane

Klio2033 [76]

Answer:

The answer to your question is below

Explanation:

Material Classification Reason

Crude oil Mixture Because we know that crude oil

is composed of different substances.

Butane Compound Because it is composed of carbon

and hydrogen

Dodecane Compound Because it is composed of carbon

and hydrogen.

Octane Compound Because it is composed of carbon and

hydrogen.

Benzene Compound It is also composed of carbon and

hydrogen.

Gasoline Mixture It is a combination of hydrocarbons

Kerosene Mixture It is a combination of hydrocarbons

Oxygen Element We can find it in the periodic table.

Carbon dioxide Compound It is composed of carbon and oxygen.

Water Compound It is composed of hydrogen and

oxygen.

8 0

4 years ago

Read 2 more answers

What is the name of this hydrocarbon?

vovikov84 [41]

A) 2,3-dimethylpentane

6 0

4 years ago

Read 2 more answers

What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative

kirill115 [55]

Answer:

HBr + H2O = H3O+ + Br-

So our conjugate acid is the H3O+ to H2O

Explanation:

A conjugate acid of a base results when the base accepts a proton.

Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:

NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.

3 0

3 years ago

A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0

4 years ago