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Vlad [161]
3 years ago
9

In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student genera

ted 3.7 g of barium carbonate from 2.0 g of their alkali metal carbonate. M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s) How many moles of alkali metal carbonate were reacted? Question 3 options: 0.019 mol 0.038 mol 0.094 mol 2.3 mol
Chemistry
1 answer:
Paraphin [41]3 years ago
3 0

Answer:

0.019 moles of M2CO3

Explanation:

M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)

From the equation above;

1 mol of  M2CO3 reacts to produce 1 mol of BaCO3

Mass of BaCO3 formed = 3.7g

Molar mass of BaCO3 = 197.34g/mol

Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187  ≈ 0.019mol

Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

1 = 1

x = 0.019

x = 0.019 moles of M2CO3

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To be able to determine the number of moles that a certain number of molecules comprises, we simply divide the number of molecules by the Avogadro's number which is equal to 6.022 x 10^23. 

   n = M/A

where n is the number of moles, M is the number of molecules, and A is Avogadro's number. Substituting the known values,

   n = (4.15 x 10^23 molecules)/(6.022 x 10^23 molecules/mol)

Simplifying,
  
   n = 0.689 moles

<em>Answer: 0.689 moles</em>

7 0
3 years ago
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What energy transfers happen when you cook sausages on a camp fire burning wood?
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3 years ago
If the density of blood is 1. 060 g/ml. What is the mass of 6. 56 pints of blood?
ryzh [129]

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7 0
2 years ago
You find a clean 100-ml beaker, label it "#1", and place it on a tared electronic balance. You add small amount of unknown solid
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Answer:

the mas is .291 g

Explanation:

the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g

7 0
3 years ago
The radioisotope radon-222 has a half-life of 3.8 days. How much of a 73.9-g sample of radon- (3 222 would be left after approxi
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Answer:

1.115 g

Explanation:

Applying,

R = R'2^{n/t}................... Equation 1

Where R = original sample of radon-222, R' = sample of radon-222 left after decay, n = Total time, t = half-life.

make R' the subject of the equation

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From the question,

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Substitute these values into equation 2

R' = 73.9/(2^{23/3.8})

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6 0
3 years ago
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