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3 years ago

How do scientists determine the number of neutrons in an isotope of an atom?

2 answers:

5 0

Scientists can calculate the number of neutrons by subtracting the atomic number from the atomic mass. The atomic number of an element is the number of protons which can be found in the nucleus of the element's atoms. The atomic mass of an element is the product of protons plus neutrons. The number of neutrons is not always equal to the number of protons. By subtracting the atomic number from the atomic mass, the difference will be how many neutrons are in the element's nucleus.

gulaghasi [49]3 years ago

5 0

D. They subtract the atomic number from the atomic mass.

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Mass is 5 grams and the volume is 4.5 grams what is the density

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Density = Mass/Volume

d= 5 / 4.5

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Which element do all organic compounds contain?

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Answer:

hydrogen oxygen or nitrogen

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What is the air pressure on top of the mountain?

mart [117]

Explanation:

At sea level, the atmospheric pressure would be a little over 100 kPa (one atmosphere or 760 mm Hg). If we climb to the top of Mount Everest (the highest mountain in the world at 29,029 feet or 8848 meters), the atmospheric pressure will drop to slightly over 30 kPa (about 0.30 atmospheres or 228 mm Hg).

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3 years ago

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the

trapecia [35]

Answer:

-21 kJ·mol⁻¹

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}$

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

$\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

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3 years ago