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AnnZ [28]
3 years ago
11

What is the name of this hydrocarbon

Chemistry
1 answer:
Bezzdna [24]3 years ago
7 0
You haven't shown the structural formula but it's either an Alkene(double bond c=b between 2 carbons) or an alkane
If there is 1 carbon it is methane (methene is not possible as a carbon cannot double bond with itself)
If there are 2 carbons it could be ethane or ethene
If there are 3 carbons it could be propane or propene
If there are 4 carbons it could be butane or butene
If there are 5 carbons it could be pentane or pentene
Etc.

E.g ethane E.g. Ethene
H H H H
| | | |
H---C---C---H C = C
| | | |
H H H H
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12.2 scientific notation
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What was the key design change for hfc-134a a/c systems versus CFC 12 a/c systems
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What type of reaction is this?—- Large Compound=Element A+Element B
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2 years ago
Balance and classify the following equations as synthesis, decomposition, single replacement, double replacement, or combustion:
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8 0
3 years ago
When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?
irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

                 AgNO₃(aq)  +  NH₄Cl(aq)   ⇒   AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of     AgNO₃(aq)  to  NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

                     1 mol of NH₄Cl  ---------------- 1 mol of AgCl

              0.00188 mol of NH₄Cl ------------- x

                     x = (0.00188 x 1) /1

                     x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

                         143.5 grams of AgCl -------------- 1 mol

                         x -------------------------------------------0.00188 moles of AgCl

                          x = (0.00188 x 143.5) / 1

                          x = 0.269 grams of AgCl

8 0
3 years ago
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