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geniusboy [140]

3 years ago

9

An uncharged atom of mercury has an atomic number of 80 and an atomic mass of 200. This atom has protons, neut

rons, and ________ electrons

1 answer:

SCORPION-xisa [38]3 years ago

3 0

<h3>Answer:</h3>

This atom has 80 protons, 120 neutrons and 80 electrons.

<h3>Explanation:</h3>

Number of Protons:

As the number of protons present in the nucleus are equal to the atomic number therefore. As given, the atomic number is 80 therefore, Hg contains 80 protons respectively.

Number of Electrons:

As given in statement, Hg is not carrying any type of charge. It means this atom is neutral in nature. Which means, it is carrying the same number of electrons as the number of protons in order to give a neutral atom. Therefore, it contains 80 electrons.

Number of Neutrons:

Number of neutrons can be calculated using following formula,

No. of Neutrons = Atomic Mass - Number of Protons

As,

Atomic Mass = 200

So,

No. of Neutrons = 200 - 80

No. of Neutrons = 120

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How many moles of oxygen are needed to react with 87 grams of aluminum

labwork [276]

Answer:

2.4 moles of oxygen are needed to react with 87 g of aluminium.

Explanation:

Chemical equation:

4Al(s) + 3O₂(l) → 2AlO₃(s)

Given data:

Mass of aluminium = 87 g

Moles of oxygen needed = ?

Solution:

Moles of aluminium:

Number of moles of aluminium= Mass/ molar mass

Number of moles of aluminium= 87 g/ 27 g/mol

Number of moles of aluminium= 3.2 mol

Now we will compare the moles of aluminium with oxygen.

Al : O₂

4 : 3

3.2 : 3/4×3.2 = 2.4 mol

2.4 moles of oxygen are needed to react with 87 g of aluminium.

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2 years ago

On a distance-time graph, at 2 hours the graph is at a height of 20 meters, and at 3 hours it is at a height of 100 meters. What

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Answer:

80 m/hr

Explanation:

Changes from 20 to 100 meters in ONE Hour

changes 80 meters in one hour = 80 m/hr

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2 years ago

The diagram below shows a cell placed in a solution.

Sholpan [36]

it will expand as water moves into it.

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2 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

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3 years ago

What is the product of the unbalanced combustion reaction below?<br> C4H10(g) + O2(g) →

Afina-wow [57]

Answer:

Option C . CO2(g) + H2O(g)

Explanation:

When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

Thus, the product of the unbalanced combustion reaction is:

CO2(g) + H2O(g)

Thus, we can balance the equation as follow:

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)

There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:

C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)

Thus, the equation is balanced.

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2 years ago