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DanielleElmas [232]
2 years ago
10

In the following acid-base reaction, NH4+ is the ____________.

Chemistry
1 answer:
Sunny_sXe [5.5K]2 years ago
4 0
Conjugate acid, I think
You might be interested in
How much heat (kJ) is absorbed by 229.1 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?
hammer [34]

Answer:

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S.I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm(t₂-t₁) ........................ Equation 1

(Q1)

Q = cm(t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991×4200(32.5-25)

Q = 1256.22(7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm(t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034×4200(23-9)

Q = 142.8(14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ

4 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
What do coefficients in chemical equations tell you?
Juliette [100K]
More information ^^^
5 0
3 years ago
List 3 elements that have similar properties or reactivity. Justify your answer with reasoning.
Rus_ich [418]

Answer:

Lithium, Potassium and Calcium

Explanation:

Because they are all in group one of the periodic table

7 0
3 years ago
In the reaction: NH4+ + H2O ⮀ NH3 + H3O+ water is acting as a(n)
Anastasy [175]

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

4 0
2 years ago
Read 2 more answers
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