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Nezavi [6.7K]
2 years ago
11

Will an atom gain or lose an electron? CER

Chemistry
1 answer:
nikdorinn [45]2 years ago
5 0
Ions are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells. When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.
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How many hydrogen atoms are in 5.20 mol of ammonium sulfide?
Fofino [41]
1) Molecular formula of ammonium sulfide

(NH4)2 S

2) That means that there are 2*4 = 8 atoms of hydrogen in each molecule of ammoium sulfide, so in 5.20 mol of molecules will be 8 * 5.20 mol = 41.6 moles of atoms of hydrogen

3) To pass to number of atoms multiply by Avogadro's number: 6.022 * 10^23

41.6 moles * 6.022 * 10^23 atoms / mol = 250.5 * 10^23 = 2.50 * 10^25 atoms

Answer: 2.50 * 10^25
8 0
3 years ago
What is the mole fraction of HCl in a solution that contains 0.99 mole HCl and 3.6 mole H2O
marysya [2.9K]

Answer:

  0.216

Explanation:

The mole fraction of HCl is the number of moles of HCl (0.99) divided by the total number of moles in the solution.

  mole fraction = (0.99)/(0.99 +3.6) = 0.2157

The mole fraction of HCl is about 0.216.

4 0
3 years ago
State three acids, bases, and Salts commonly used in therapeutic processes​
Delvig [45]
Acetic acid, phosphoric acid, hydrochloric acid.

sodium bicarbonate, magnesium hydroxide, aluminum hydroxide

calcium carbonate, sodium bicarbonate, magnesium sulfate.
3 0
2 years ago
How many pounds of coal would you have to burn to generate enough heat to boil a 15 l kettle?
rjkz [21]
Note that
The heating value of standard coal is about 30,080 kJ/kg
1 L of water has a mass of 1.0 kg

The mass of 15 L of water = 15 kg.
The latent heat of vaporization of water is about 2260 kJ/kg,
 The energy required to boil 15 L of water is
(2260 kJ/kg)*(15 kg) = 33900 kJ

The mass of coal required to provide this energy is
(33900 kJ)/(30080 kJ/kg) = 1.127 kg

Because 1 kg = 2.205 lb, the mass of coal required is
(1.127 kg)*(2.205 lb/kg) = 2.485 lb

Answer: 2.49 lb (nearest hundredth)

7 0
2 years ago
An insulin drip is mixed as 500.0 units in 250.0 mL Normal Saline (NS). Calculate the drip rate in mL/h needed to deliver 4.0 un
olasank [31]

Answer:

Drip rate = 2 ml/h

Explanation:

Given:

Amount of insulin = 500 unit

Amount of saline = 250 ml

Required infusion rate = 4.0 units/h

Find:

Drip rate

Computation:

Concentration = Units of insulin / Volume of saline

= 500 / 250

Concentration = 2 units/ml

Drip rate = Infusion rate / concentration

Drip rate = 4 / 2

Drip rate = 2 ml/h

3 0
2 years ago
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