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2 years ago

Which property of fossil fuel make them an important resource for life on earth

Chemistry

1 answer:

grandymaker [24]2 years ago

7 0

Fossil Fuels give off energy when they are burned

One valuable fossil fuel is natural gas. It is a cleaner-burning fuel source.

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Provide the alkyne starting material a, intermediate organic product b, and identify the reagents.

Oksi-84 [34.3K]

Co2 is correct buddy

7 0

2 years ago

Need help with this please help ASAP

gregori [183]

Answer:

The correct answer would be -

17. 50% chances of round offspring

18. 25% chances two copies of dominant allele in the genotype of offspring

Explanation:

17. In this question it is stated that the Spongebob is heterozygous and has a square body which means the square body is dominant over the round that is present in sponge Susie that means she has two copies of the recessive allele.

S for square trait and s for a round trait then

S s

s Ss ss

Thus, the correct answer is 50 percent as there are only two copies of ss.

18. in this question Patric and Patti both have heterozygous for the pink (P) body-color dominant over yellow (p). The chances for the true-breeding dominant genotype would be -

P p

P PP Pp

p Pp pp

There is only one out of four rue-breeding dominant offspring will be produced.

3 0

2 years ago

Calculate the number of moles in 23.8 grams of Nitrogen, N.

Vlad1618 [11]

Answer:

n = 1.7 moles

Explanation:

Given mass, m = 23.8 grams

Molar mass of Nitrogen = 14 g/mol

Let there is n number of moles. We know that,

No. of moles = given mass/ molar mass

$n=\dfrac{23.8}{14}\\\\n=1.7$

So, there are 1.7 moles in 23.8 grams of Nitrogen.

8 0

2 years ago

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: $V_{1}$ = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

$M_{1}$ = 0.321 M, $V_{2}$ = 21.8 mL = 0.0218 L, $M_{2}$ = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$

Substitute the values into above formula as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M$