All categories

2 years ago

The table shows the number of charged subatomic particles in an ion.

Chemistry

2 answers:

LenKa [72]2 years ago

7 0

Answer:

The ion will repel the substance because it has more protons than electrons.

Explanation:

Rzqust [24]2 years ago

5 0

Answer: Option (c) is the correct answer.

Explanation:

It is known that sub-atomic particles present in an atom are protons, neutrons and electrons.

It is given that there are 2 positive sub-atomic particles present in the given atom. This means there are 2 protons.

And, it is give that there are 3 negative sub-atomic particles. This means that there are 3 electrons.

Also, when an atom is neutral then there are same number of protons and electrons.

This also shows that the given ion contains a negative charge. As like charges repel each other so, when a negatively charged substance is brought near the ion then there will occur repulsion between them.

Thus, we can conclude that the ion will repel the substance because it has more electrons than protons.

You might be interested in

How many lead atoms are in a cube of lead that has a volume of 2.18 cm3 if the density of lead is 11.3 g/cm3?

Anit [1.1K]

10000000.................

3 0

3 years ago

How many salt and water would you use to prepare 10% salt solution?

tatuchka [14]

Answer:

We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.

Explanation:

have a good day

4 0

2 years ago

How many grams of acetylene are produced by adding water to 5.00 g CaC2?

Murljashka [212]

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

C = 12g/mol

H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

5g would produce x?

64 = 26

5 = x

Upon solving for x we have;

x = (5 * 26) / 64

x = 2.03125g

5 0

3 years ago

What is the molarity (molar concentration, unit = M) of K+ found in 200 mL 0.2 M K2HPO4 solution?

enyata [817]

Answer:

0.4 M

Explanation:

The process that takes place in an aqueous K₂HPO₄ solution is:

K₂HPO₄ → 2K⁺ + HPO₄⁻²

First we calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution:

200 mL * 0.2 M = 40 mmol K₂HPO₄

Then we convert K₂HPO₄ moles into K⁺ moles, using the stoichiometric coefficients of the reaction above:

40 mmol K₂HPO₄ * $\frac{2mmolK^+}{1mmolK_2HPO_4}$ = 80 mmol K⁺

Finally we divide the number of K⁺ moles by the volume, to calculate the molarity:

80 mmol K⁺ / 200 mL = 0.4 M

5 0

3 years ago

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=5.05

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

3 years ago