Ask question

Ask question

All categories

4 years ago

15

True or False, A Bronsted-Lowry acid is a proton acceptor, while a Bronsted-Lowry base is a proton donor.

1 answer:

melomori [17]4 years ago

8 0

This statement is false. A Bronsted-Lowry acid is a proton DONOR and the Bronsted-Lowry base is a proton ACCEPTOR, Based on the Bronsted-Lowry definition, acids and bases are being classified whether they can donate or accept and proton. Bronsted-Lowry acid is a PROTON DONOR because it donates a proton to H2O and for the Bronsted-Lowry base, it basically accepts a proton from acids such as HCl.

You might be interested in

Subtracting atomic number from atomic mass will give the number of

faltersainse [42]

Answer:

:)

Explanation:

If we subtract the atomic number from the atomic mass: atomic mass - atomic number = number of protons + number of neutrons - number of protons. Thus we get the number of neutrons present in an atom when we subtract the atomic number from the atomic mass.

8 0

3 years ago

Read 2 more answers

Pls help i will give crown

Gelneren [198K]

Answer: decay

Explanation:

7 0

3 years ago

Read 2 more answers

Who can help me please ? Yes I will give BRAINLIST . And yes this is 6th grade work . Please help me ( it’s science btw )

lana [24]

Answer:

true and b are the answers I think

6 0

3 years ago

Read 2 more answers

Combustion of hydrocarbons such as pentane (C5H12) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's

zhuklara [117]

2 C₅H₁₀ (l) + 15 O₂ (g) → 10 CO₂ (g) + 10 H₂O (g)

Explanation:

Balanced chemical equation for combustion of pentane C₅H₁₀:

C₅H₁₀ (l) + (15/2) O₂ (g) → 5 CO₂ (g) + 5 H₂O (g)

to get integer numbers for the stoechiometric coefficients we multiply with 2:

2 C₅H₁₀ (l) + 15 O₂ (g) → 10 CO₂ (g) + 10 H₂O (g)

where:

(l) - liquid

(g) - gas

Learn more about:

combustion of hydrocarbons

brainly.com/question/13691565

#learnwithBrainly

5 0

3 years ago

Read 2 more answers

Uranium-232 has a half-life of 68. 8 years. After 344. 0 years, how much uranium-232 will remain from a 100. 0-g sample? 1. 56 g

Ahat [919]

The remaining mass of Uranium -232 has been 3.13 g. Thus, option B is correct.

The half life has been defined as the time required by the substance to reduce to half of its initial value.

The half life has been expressed as:

$N_t=N_0\;(\dfrac{1}{2})^\frac{t}{t_{1/2}}$

<h3>Computation for the remaining Uranium-232</h3>

The initial mass of uranium, $N_0=100\;\rm g$

The half-life of uranium has been, $t_\frac{1}{2} =68.8\;\rm years$

The time required for the reducing has been, $t=344\;\rm years$

Substituting the values for the remaining mass of Uranium-232 ( $N_t$ ):

$N_t=100\;\times\;(\dfrac{1}{2})^\frac{344}{68.8} \\
N_t=100\;\times\;(\dfrac{1}{2} )^5\\
N_t=3.13\;\rm g$

The remaining mass of Uranium -232 has been 3.13 g. Thus, option B is correct.

Learn more about half-life, here:

brainly.com/question/24710827

8 0

2 years ago