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4 years ago

Provide the Bronstead-Lowry de finition for an "Acid" and a "Base"

Chemistry

1 answer:

Setler79 [48]4 years ago

3 0

Answer:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For example:

$H_3BO_3(aq.)+HS^-(aq.)\rightarrow H_2BO_3^-(aq.)+H_2S(aq.)$

Here, $H_3BO_3$ is loosing a proton, thus it is considered as a bronsted acid and after losing a proton, it forms $H_2BO_3^-$ which is a conjugate base.

And, $HS^-$ is gaining a proton, thus it is considered as a bronsted base and after gaining a proton, it forms $H_2S$ which is a conjugate acid.

The sequence of the acids and bases in the above equation are:

$Acid+Base\rightarrow Base+Acid$

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A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measure

Leni [432]

Answer: $(27.81,\ 29.39)$

Explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level : $\alpha: 1-0.95=0.05$

Critical value: $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}=28.6$

Standard deviation : $\sigma=2.2$

The formula to find the confidence interval is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}$

i.e. $28.6\pm 0.787259889321$

$\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)$

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = $(27.81,\ 29.39)$

4 0

3 years ago

Covalent bonds take place between..

umka2103 [35]

Answer:

Covalent Bonds are formed when two non-metals share electrons

Hope this helps

7 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

3 years ago

What are some properties of matter that can be identified without testing or measuring

kiruha [24]

Physical change - No change of matter in this phase
chemical change - All types of phase change occur here

8 0

3 years ago

V = d/t

dangina [55]

Answer:

Explanation:

Did you mean: V = d/t a = (V - Vit Average = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?

Showing results for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = "-9.81" m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?

Search instead for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?

7 0

3 years ago