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dem82 [27]
3 years ago
9

Which type of molecule is shown below?

Chemistry
2 answers:
zubka84 [21]3 years ago
8 0

Answer:

\boxed{\mathrm{Alkene}}

Explanation:

The hydrocarbon shown has a double bond. Hydrocarbons with double bonds are known as alkenes.

Cyclic alkanes have cyclic structure.

Alkanes only have single bonds.

Alkynes have triple bonds.

GenaCL600 [577]3 years ago
7 0

Answer:

Alkene

Explanation:

A molecule having a double bond between its atoms is called an alkene.

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What is the pH of a solution when the [OH - ] = 1 x 10 -3
dsp73

Answer:

The pH is 11

Explanation:

Because pH + pOH = 14

and pOH = - log [OH⁻] = - log (1 x 10⁻³) = 3

we can now calculate pH by manipulating equation one above

pH = 14 - pOH = 14 - 3 = 11

Thus the pH of the solution is 11

8 0
3 years ago
What is the percentage composition of each element in hydrogen peroxide, H2O2?
Degger [83]

Answer:

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

Explanation:

Step 1: Data given

Molar mass of H = 1.0 g/mol

Molar mass of O = 16 g/mol

Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol

Step 2: Calculate % hydrogen

% Hydrogen = ((2*1.0) / 34.0) * 100 %

% hydrogen = 5.88 %

Step 3: Calculate % oxygen

% Oxygen = ((2*16)/34)

% oxygen = 94.12 %

We can control this by the following equation

100 % - 5.88 % = 94.12 %

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

8 0
3 years ago
Read 2 more answers
How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?
melisa1 [442]

When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.

- Heat given out by water = Heat absorbed by ice

Calculating the heat released by hot water:

q = m CwaterΔTq = m (4.184\frac{J}{(g.^{0}C)} )(0^{0}C-50^{0}C)

Calculating heat absorbed by 16 g of ice: Ice at -12^{0}C is converted to ice at O^{0}C and then ice at O^{0} C to water at 0^{0}C

q = m CiceΔT + m (Heat of fusion)

q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C)) + 16 g (333.55\frac{J}{g})

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-(m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J

m = 27.4 g

Therefore, 27.4 g water at 50^{0}C must be added to 16 g of ice at -12^{0}C to convert to liquid water at 0^{0}C

4 0
3 years ago
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Answer:

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Explanation:

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