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dlinn [17]
3 years ago
15

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How

many carbon atoms are in 1.50 g of butane?
Chemistry
1 answer:
forsale [732]3 years ago
7 0
Doesnt the number of carbon atoms stay the same.
Though the weight of carbon in 1.5g is 1.24g.

This is because the RAM of C4 is 48.

The RFM of C4H10 is 58. Therefore, 48/58 is carbon in butane.

48/58 x 1.5 = 1.24g
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A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r
Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added =  0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 ×  0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 =  0.005 moles -  0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3  in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

5 0
2 years ago
Please help with this question
tamaranim1 [39]
1.806x10^24
Written equation form(always start the equation off with what you know based off of the question!):

3mol(CCl4)•6.022x10^23/1mol = 1.806x10^24

Good luck!
3 0
3 years ago
2HCl + CaCO3 → CaCl2 + H2O + CO2
zloy xaker [14]

Answer:

The amount of CaCl2 produced depends on the amount of HCl in the reaction.

Explanation:

The amount of HCl is used completelyin the reaction unlike CaCO3 which remains after reaction.

4 0
3 years ago
Why is there a difference in the conductivity of pure solid NaC1 and of the 1.0 M NaC1 soulution
Basile [38]
That would be cause part of the sodium is pure and that means it still kind of has it properties when it was an element and that its i think.
7 0
3 years ago
100 POINTS PLEASE HELP (PHOTOS INCLUDED)
pshichka [43]

Answer:

Explanation:

3.

Knowns: 100mL of solution; concentration of 0.7M

Unknown: number of moles

Equation: number of moles = volume * concentration

Plug and Chug: number of moles = 100/1000 * 0.7 = 0.07 mole

Final Answer: 0.07mole

2.

Knowns: 5.50L of solution; concentration of 0.400M

Unknown: number of moles

Equation: number of moles = volume * concentration

Plug and Chug: number of moles = 5.5 * 0.4 = 2.20 mole

Final Answer: 2.20 mole

6 0
2 years ago
Read 2 more answers
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