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drek231 [11]
3 years ago
9

A 0.75mol calcium carbonate (CaCO3) pellet has what mass?

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
4 0

Answer:

75 g  

Step-by-step explanation:

<em>Step 1. </em>Calculate the molar mass of CaCO₃

Ca                    =   40.08

  C                    =   12.01

3O = 3 × 16.00 = <u>  48.00 </u>

               Total = 100.09 g/mol

<em>Step 2</em>. Calculate the mass of the pellet

Mass of CaCO₃ = 0.75 mol × (100.09 g/1 mol)

Mass of CaCO₃ = 75 g

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Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li3N(s) + 3H2O (L) --&gt;
Gemiola [76]

Answer:

0.480 grams

Explanation:

Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)

1             :  3                                            : 1           :  3

Number of moles (n) = Mass in gram/ Molar Mass

Mass of ND₃ = 160 mg

                     = 0.16 g

Molar mass of ND₃=  [14 + (3 x 2.014 )]

                               =    14 + 6.042

                               =  20.042 g/mol

Number of moles of   ND₃  =  0.16/20.042

                                             =  0.007983 moles

From the reaction equation, the mole ratio between   Heavy water (D₂O ) and  ND₃ is  3: 1.

This implies that the number of moles of   Heavy water (D₂O )  required

= 3  x 0.007983 moles

=  0.023949 moles

Molar mass of  Heavy water (D₂O )=  [(2.014 x 2) + 16]

                                                           =  20.028 g/mol

Mass in grams of Heavy water (D₂O )= Number of moles  x Molar mass

                                                            =   0.023949   x  20.028

                                                            =  0.4797 grams

                                                            ≈ 0.480 grams

3 0
3 years ago
In the chemical reaction:
velikii [3]

Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Cu + 2 AgNO₃  → 2 Ag + Cu(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Cu: 1 mole
  • AgNO₃: 2 moles
  • Ag: 2 moles
  • Cu(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?

amount of moles of AgNO_{3} =\frac{1.8 moles of Cux2 moles of AgNO_{3} }{1 mole of Cu}

<u><em>amount of moles of AgNO₃= 3.6 moles</em></u>

But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.

<h3>Summary</h3>

In summary, AgNO₃ will be the limiting reagent.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

8 0
1 year ago
Calculate the grams of CaCl2 necessary to make a 0.15Msolution.
xxTIMURxx [149]

Answer: mass m = M·c·V

Explanation: M(CaCl2) = 110.98 g/mol, c= 0.15 mol/l,

n=m/M= cV, volume of Solution is not mentioned

3 0
3 years ago
the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107
oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u)  and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0
3 years ago
How many moles of oxygen atoms do 1.5 moles of co2 contain?
IgorLugansk [536]
1 molecule CO2 has 2 atoms O.
1 mole CO2 has 2 moles O,
1.5 mole CO2 has 2*1.5 mole O=3.0 mole O
7 0
4 years ago
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