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Mandarinka [93]

3 years ago

14

Consider the unbalanced equation for the oxidation of butene. C4H8 + 6O2 Right arrow. CO2 + H2O For each molecule of C4H8 that r

eacts, how many molecules of carbon dioxide and water are produced? Group of answer choices 2 CO2 & 2 H2O 4 CO2 & 4 H2O 2 CO2 & 4 H2O 4 CO2 & 2 H2O

1 answer:

antiseptic1488 [7]3 years ago

4 0

Answer is C4H8 + 6O2 —> 4CO2 + 4H2O

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How would you prepare 175 mL of a 0.350kmol/m^3 calcium nitrate solution

SVEN [57.7K]

0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol

mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol

1 mol --------- 164g
0,06125 ---- X
X = 10,045g

To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.

3 0

3 years ago

URGENT!! Can someone please explain this to me?

djverab [1.8K]

Answer:

$CH _{4}+ 2O _{2} → CO _{2} + 2H _{2}O$

for the balanced equation

$from \: the \: equation \\ 1 \: mole \: of \: methane \: gives \: 1 \: mole \: of \: carbon \: dioxide \\ 7.4 \: moles \: of \: carbon \: dioxide \: will \: be \: given \: by \: (7.4 \times 1)moles \: of \: methane \\ = 7.4\: moles \: of \: methane \\ \\ since \: moles \: of \: oxygen \: double \: those \: of \: methane \\ moles \: of \: oxygen \: = 7.4\times 2 \\ = 14.8 \: moles \: of \: oxygen$

4 0

3 years ago

Number of molecules in 4HCl

katrin2010 [14]

Answer:

its not cear to answer this question

6 0

3 years ago

Element 2 is:<br> A) Cobalt<br> B)Chlorine <br> C)Calcium<br> D)Carbon

saul85 [17]

Answer:

D: Carbon

Explanation:

Carbon is the sixth element with a total of 6 electrons in the periodic table. Hence the atomic number Z = 6. The ground state electron configuration of carbon is 1s2 2s2 2p2. An excited state electron configuration of carbon is 1s2 2s1 2p3.

4 0

3 years ago

What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C

Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = $\frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\$

Now use the molar ratio to find how many moles of NaOH would be required to neutralize $5.5537*10^{-3} mol\\$ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

1 : 2 (we get this from the balanced equation)

$5.5537*10^{-3} mol\\$ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

$c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L$

5 0

4 years ago