An acid is a chemical which "wants" to donate some protons, or hydrogen + ions. Since a hydrogen atom is just a proton and an electron, the ion lacking an electron is simply a proton. Hope this helps!

6 0

3 years ago

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1. How would the series of figures change in the presence of a catalyst?

statuscvo [17]

No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.

hope this explains it

If it does, please give it a brainliest :)))

8 0

2 years ago

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What is the empirical formula of a compound with a percent composition of 22.5% Phosphorous and 77.5% Chlorine?

sashaice [31]

Answer:

$\boxed {\boxed {\sf PCl_3}}$

Explanation:

We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.

We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.

22.5 g P
77.5 g Cl

Next, convert these masses to moles, using the molar masses found on the Periodic Table.

P: 30.974 g/mol
Cl: 35.45 g/mol

Use the molar masses as ratios and multiply by the number of grams. $22.5 \ g \ P * \frac {1 \ mol \ P }{30.974 \ g \ P}= \frac {22.5 \ mol \ P }{ 30.974} = 0.7264157035 \ mol \ P$

$77.5 \ g \ Cl * \frac {1 \ mol \ Cl }{35.45 \ g \ Cl}= \frac {77.5 \ mol \ Cl }{ 35.45} \ =2.186177715 \ mol \ Cl$

Divide both of the moles by the smallest number of moles to find the mole ratio.

$\frac {0.7264157035} {0.7264157035} = 1$

$\frac {2.186177715}{0.7264157035}=3.009540824 \approx 3$

The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as:<u> PCl₃</u>

4 0

2 years ago