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3 years ago

An atom has a mass number of 24 and 13 neutrons. What is the atomic number of this atom?

2 answers:

6 0

A=Mass number=24
N=neutrons=13
Z=atomic number.

A=Z+N

24=Z+13
Z=24-13
Z=11

The atomic number is 11, and this atom is sodium.

Lady_Fox [76]3 years ago

3 0

11,the numbers of protons and electrons are always equal in an atom,the numbers of protons tell us about atomic number of element
mass number=no. of P+no. of N
here,24
no. of P(atomic number)=24-13=11

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Glycerol (molar mass 92.09 g/mol) has been suggested for use as an alternative fuel. The enthalpy of combustion, Δ

victus00 [196]

The mass of glycerol to that would need to be combusted to heat 500.0g of water from 20.0°C to 100.0°C is; 9.32 grams.

We must establish the fact that energy is neither created nor destroyed.

Therefore, the amount of heat absorbed by water is equal to the amount of heat released by the combustion of glycerol.

Total heat absorbed by water, H(water) is;

Q(water) = m C (T2 - T1)

Q(water) = 500 × 4.184 × (100-20)

Q(water) = 167.36 kJ

Consequently, the quantity of heat evolved by the combustion of glycerol is;

Q(glycerol) = 167,360 J = n × ΔH°comb

where, n = no. of moles of glycerol.

167.36 kJ= n × 1654 kJ/mole

n = 167.36/1654

n = 0.1012 moles of glycerol.

Therefore, mass of glycerol combusted, m is;

m = n × Molar mass

m = 0.1012 × 92.09

m = 9.32 g.

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2 years ago

Consider the following intermediate chemical equations.

QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$ $\Delta H_1=-2439kJ$ (1)

$4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)$ $\Delta H_2=3438kJ$ (2)

Net chemical equation:

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H=?$ (3)

Adding 1 and 2 we get,

$4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2$ $\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ$ (4)

Now dividing equation (4) by 4, we get

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2$

$\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ$ (4)

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2 years ago

Solid lead IV oxide decomposes into solid lead and oxygen gas.

vredina [299]

The balanced equation for the decomposition of solid lead iv oxide is as follows: 2PbO2 = 2PbO + O2.
Lead IV oxide decompose to give lead ll oxide and oxygen. Lead iv oxide is thermally unstable and it usually decomposes into oxygen and lead ll oxide when heated. Lead ll oxide is more stable than lead lV oxide.

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3 years ago

The elements in which electrons are progressively filled in 4f-orbital are called.

Y_Kistochka [10]

The elements in which electrons are progressively filled in 4f-orbital are called lanthanoids.

There are 15 metallic chemical elements called lanthanoids.

Those elements are with atomic numbers from 57 to 71.

Lanthanoids belong to f-block elements because of the filling of the 4f-orbitals.

One of the lanthanoids, lutetium (chemical symbol Lu) has 14 electrons if 4f orbitals.

Electron configuration of lutetium: [Xe] 4f14 5d1 6s2

f-orbitals are similar to the d orbital, but cut in half (eight lobes instead of four).

Lanthanoids are used not as materials in catalysts, alloys, lasers and cathode-ray tubes.

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aleksklad [387]

Answer:

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3 years ago