An element has atomic number 10 and an atomic mass of 20. How many neutrons are in the atom of this element?

1 answer:

8 0

Atomic mass is found by adding the number of protons and neutrons together. The atomic number tells you how many protons an element has.
So since this element has 10 protons and an atomic mass of 20, there are 10 neutrons in this element.

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6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?

Darina [25.2K]

Copper heat capacity would be 0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= 0.385J/C*gram * 6g * (90-20)
energy= 161.7J

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3 years ago

Why is it important for scientists to use the scientific method?

MakcuM [25]

Explanation:

It provides an objective, standardized approach to conducting experiments and, in doing so, improves their results. By using a standardized approach in their investigations, scientists can feel confident that they will stick to the facts and limit the influence of personal, preconceived notions.

I hope this helps you out!

4 0

2 years ago

Read 2 more answers

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$