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attashe74 [19]
2 years ago
15

What is the molar mass of Ca(NO3)2

Chemistry
2 answers:
artcher [175]2 years ago
7 0
Ca=40
N=14
O=16
Ca=40
N=14x2=28
O=6x16=96
molar mass=40+28+96
molar mass=164g/mol

AleksAgata [21]2 years ago
4 0

Answer:

Molar mass = 164 grams per mol

Explanation:

The compound above is calcium nitrate represented as Ca(NO3)2. The molar mass of the compound can be calculated from the atomic mass of each atom present in the compound.

Atomic mass of calcium = 40 g

Atomic mass of nitrogen  = 14 g

atomic mass of oxygen = 16 g

Ca is 1 mole = 40 grams

Nitrogen is 2 moles = 14 × 2 = 28 g

oxygen is 6 moles = 16 × 6 = 96 g

The sum = 40 + 28 + 96 = 164 grams per mol

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If 1.2 moles of ideal gas occupy a volume of 18.2 L at a pressure of 1.8 ATM what is the temperature of gas in degrees Celsius.
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Answer: A 59.5 degree celcius

The equation that we will use to solve this problem is :

PV = nRT where:

P is the pressure of gas = 1.8 atm

V is the volume of gas = 18.2 liters

n is the number of moles of gas = 1.2 moles

R is the gas constant =  0.0821

T is the temperature required (calculated in kelvin)

Using these values to substitute in the equation, we find that:

(1.8)(18.2) = (1.2)(0.0821)(T)

T = 332.5 degree kelvin

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T = 332.5 - 273 = 59.5 degree celcius

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3 years ago
Rank these elements according to FIRST ionization energy, from highest to lowest:
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Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following
bearhunter [10]

Answer:

210.7~g~MgCO_3

Explanation:

We have to start with the <u>reaction</u>:

MgCO_3~->~MgO~+~CO_2

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

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If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

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In other words: 1~mol~CO_2=~44~g~CO_2. With this in mind, we can calculate the moles:

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Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

2.5~mol~CO_2=2.5~mol~MgCO_3

With the molar mass of MgCO_3 ((23.3*1)+(12*1)+(16*3)=84.3~g/mol. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3

I hope it helps!

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3 years ago
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