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Paladinen [302]
3 years ago
11

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

Chemistry
1 answer:
kirill115 [55]3 years ago
3 0

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.

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If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
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<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

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