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3 years ago

When a substance accepts a proton (H+) what is it becoming? Is it becoming a negative or positive ?

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

6 0

Answer:

It’s becoming more positive, although it doesn’t inherently make it positive

Explanation:

Was the question worded like this? If so then the answer is def positive

You might be interested in

hich of these statements is supported by the results of Thomson’s experiment? Check all of the boxes that apply.

snow_tiger [21]

Cathode rays are made up of negatively charged particles.

Atoms contain negatively charged particles.

Explanation:

The experiment carried out by J.J Thomson provided evidence to support that fact that atoms contains negatively charged particles and that cathode rays are negatively charged.

J.J Thomson performed experiments using a gas discharge tube.

Through his experiment, he discovered cathode rays.
From his observation, he noticed that they were attracted to the another so they must be negatively charged.
He called the rays cathode rays.
Today, they are known as electrons.
Thomson's experiment discarded the widely held believed that atoms are indivisible.

learn more:

Rutherford experiment brainly.com/question/1859083

#learnwithBrainly

6 0

3 years ago

Read 2 more answers

Need help !!!!! ASAP

Nat2105 [25]

<h2>Hello!</h2>

The answer is:

The temperature will be the same, 37°C.

Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.

The law establishes the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

Then, we are given the following information:

$V_{1}=200mL\\P_{1}=4atm\\T_{1}=37\°C\\V_{2}=400mL\\P_{2}=2atm$

So, isolating the new temperature and substituting the given information, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\T_{2}=P_{2}V{2}*\frac{T_{1}}{P_{1}V_{1}} \\\\T_{2}=2atm*400mL*\frac{37\°C}{4atm*200mL}=37\°C$

Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.

The temperature will be the same, 37°C

Have a nice day!

4 0

3 years ago

Does anyone know what minerals were in the identifying a mineral project or how to fix this problem?

nadezda [96]

Answer: are you in times4learning?

Explanation:

8 0

2 years ago

Which element has four energy levels? Na,K,Cs,Rb

Ann [662]