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3 years ago

Are they all correct?

Chemistry

2 answers:

Degger [83]3 years ago

7 0

Answer:yes

Explanation:

lapo4ka [179]3 years ago

4 0

Answer:

yessirrr

Explanation:

idk about 60 tho

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Some students have said that a buffer is like a proton (H ) sponge. Evaluate this statement and explain both how a buffer is lik

alisha [4.7K]

Answer:

i) Buffer absorbs H^+ and OH^- ions preventing large changes in pH when small amounts of acid or base is added but when large amounts of acid or base is added there will be a change in pH

ii) absorption of liquid is related to a chemical reaction for a buffer system but it is not related a chemical reaction for a proton( H ) sponge.

Explanation:

i) Buffer like a proton ( H ) sponge

Buffer absorbs H^+ and OH^- ions preventing large changes in pH when small amounts of acid or base is added but when large amounts of acid or base is added there will be a change in pH

The pH of a Buffer follows the Henderson-Hasselbach model

pH = pKa + Log ([A-]/[HA])

when base is added

HA + OH^- -------> A- + H2O

when acid is added

A^- + H^+ ---------> HA

ii) Buffer not like a proton ( H ) sponge

absorption of liquid is related to a chemical reaction for a buffer system but it is not related a chemical reaction for a proton( H ) sponge.

8 0

3 years ago

1. Which statement correctly describes most cinder cone volcanoes?

Mashcka [7]

Answer:

1 a

2 d

Explanation:

5 0

4 years ago

Read 2 more answers

The speed of a car describes

Kamila [148]

It’ describes how fast the car is going

4 0

4 years ago

The half-life of iodine-131 is about 8 days. How much of a 50mg sample will be left in 25 days? Write your answer rounded to the

ASHA 777 [7]

Answer:

5.73 mg of the sample will be left in 25 days.

Explanation:

Given that:

Half life = 8 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{8}\ days^{-1}$

The rate constant, k = 0.08664 days⁻¹

Time = 25 days

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration = 50 mg

So,

$[A_t]=50\times e^{-0.08664\times 25}\ mg$

$[A_t]=5.73\ mg$

5.73 mg of the sample will be left in 25 days.

6 0

3 years ago

How many molecules (not moles) of NH3 are produced from 5.01×10−4 g of H2?

rodikova [14]

Find the number of molecules NH3 produce from the given equation:
7.42 x 10 – 4g H2
Now, let’s start solving this equation:
=> H2 reacts to ammonia and produced 3 to 2
=> Avogrado’s number = 6.02 x10^23
=> 7.42 x 10^/4 H2 times
=> 2.02g H2/1 mol H2 times
=> 3 mols H2/2 mols NH3 times
=> 6.02 x 10^23 molecules NH3/1 mol NH3
=> Thus the answer is 1.35 x 10^23 molecules NH3

6 0

4 years ago