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3 years ago

How do i know how many neutrons of an element

Chemistry

2 answers:

Alex_Xolod [135]3 years ago

8 0

Answer:

neutrons = atomic mass - atomic number.

hope that helps

alekssr [168]3 years ago

4 0

Answer:

Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).

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What is the purpose of the prefixes in the metric system?

Helen [10]

<span>Prefixes are used in the metric system to indicate smaller or larger measurements</span>

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3 years ago

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If evaporation occurs what will change 0.70 molar solution of CuSO4

faust18 [17]

Answer:

A. Molarity will increase .

Explanation:

Molarity = moles of solute per litre of solution

= moles of solute / volume of solution

If evaporation occurs , volume of solution decreases and moles of solute remains constant . Hence denominator decreases and numerator remains constant .

Hence the molarity increases .

3 0

3 years ago

A given mass of air has a volume of 6.00 L at 101 kPa. At constant temperature, the pressure is decreased to 25.0 kPa. Calculate

GenaCL600 [577]

Answer:

24.24 L

Explanation:

Boyle’s law, also called Mariotte’s law, a relation concerning the compression and expansion of a gas at constant temperature.

This empirical relation, formulated by the physicist Robert Boyle in 1662, states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form, pv = k, a constant.

Real gases obey Boyle’s law at sufficiently low pressures, although the product pv generally decreases slightly at higher pressures, where the gas begins to depart from ideal behaviour.

As, PV = k

P₁ V₁ = P₂ V₂

Given P₁ = 101 KPa

V₁ = 6 L

P₂ = 25 kPa

So, V₂ = P₁ V₁ /P₂ = 101 *6/25 = 24.24 L

4 0

3 years ago

How many grams of nitrogen dioxide are required to produce 5.89x10^3 kg of hno3 in excess water

polet [3.4K]

5 0

3 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water

S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0

3 years ago