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denpristay [2]
3 years ago
8

Chlorine has an atomic number of 17.

Chemistry
2 answers:
zlopas [31]3 years ago
6 0

Answer:

A. 17

Explanation:

pishuonlain [190]3 years ago
4 0

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Atomic number is defined as the number of protons or number of electrons that are present in a neutral atom.  It is represented by the symbol 'Z'

Z = Atomic number = number of protons = number of electrons

We are given:

Atomic number of chlorine = 17

So, number of protons = 17

Hence, the correct answer is Option A.

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A solution is prepared by adding 0.0231moles of H3O+ ions to 3.33L of water. What is the pH of this solution
MrMuchimi

Answer:

2.15

Explanation:

For this question, we have to remember the <u>pH formula</u>:

pH~=~-Log[H_3O^+]

By definition, the pH value is calculated when we do the -Log of the concentration of the <u>hydronium ions</u> (H_3O^+). So, the next step is the calculation of the <u>concentration</u> of the hydronium ions. For this, we have to use the <u>molarity formula</u>:

M=\frac{mol}{L}

We already know the number of moles (0.0231 moles) and the volume (3.33 L). So, we can plug the values into the molarity formula:

M=\frac{0.0231~moles}{3.33~L}=0.00693~M

With this value, now we can calculate the pH value:

pH~=~-Log[0.00693~M]~=~2.15

<u>The pH would be 2.15</u>

I hope it helps!

8 0
3 years ago
0.00000000082 -<br> scientific notation
Nadusha1986 [10]

Answer:

8.2 x 106^-11

Explanation:

To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2

Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.

Once the decimal is in the right spot count the spots moved.

Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.

6 0
3 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
If 3.289 x 10^23 atoms of potassium react with excess water, how many grams of hydrogen gas would be produced?
GarryVolchara [31]

Answer:

The amount in grams of hydrogen gas produced is 0.551 grams

Explanation:

The parameters given are;

Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms

Chemical equation for the reaction is given as follows;

2K + 2H₂O \rightarrow KOH + H₂

Avogadro's number, N_A, regarding the number of molecules or atom per mole is given s follows;

N_A = 6.02 × 10²³ atoms/mole

Therefore;

The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles

2 moles of potassium produces one mole of hydrogen gas, therefore;

1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas

The molar mass of hydrogen gas = 2.016 grams

Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.

The amount in grams of hydrogen gas produced = 0.551 grams.

8 0
3 years ago
HURRY I"M BEING TIMED!!!! if an unknown solution has a pH of 2. how would you classify this solution?
Sergeu [11.5K]
An acid, acids have a pH less than 7
7 0
3 years ago
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