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Aliun [14]
3 years ago
6

Which of the following would not have the same number of valence electrons as the others?

Chemistry
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

Carbon

Explanation:

Carbon has 4 valence electrons while the other ones have 6.

I appreciate it if you could mark me as Brainliest!

Fantom [35]3 years ago
5 0

carbon won't have same number of valence electrons as others

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For a lake in Michigan, researchers have determined that largemouth bass feed on smaller fish, which in turn feed on zooplankton
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Answer:

the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

Explanation:

Given the data in the question;

In 2016, there was 600,000 kg of zooplankton in the lake.

In 2017, an accidental runoff of insecticide near the lake caused a 50 percent decline of the zooplankton population in the lake.

Now,

The remaining mass of zooplankton  after the 50% decline will be;

⇒ 600,000 kg zooplankton  × 50%

⇒ 600,000 × 50/100

⇒ 300,000 kg of zooplankton

Now, with 10 percent trophic efficiency;  smaller fish directly feed on zooplankton; decline in smaller fish mass will be;

⇒ 300,000 kg × 10%

⇒ 300,000 × 10/100

⇒  30000 kg

Finally, with 10 percent trophic efficiency, largemouth bash directly feed on smaller fish, so the expected decline in mass of largemouth bash will be;

⇒ 30000 kg × 10%

⇒ 30000 kg × 10/100

⇒ 3,000 kg

Therefore, the expected decline in largemouth bass is 3,000 kg.

Option d) 3,000 kg is the correct answer.

7 0
3 years ago
How many molecules are in a mole of H2O
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Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -
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Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)  \Delta H=-2220.1kJ/mol

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]

where,

n = number of moles

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Now put all the given values in this expression, we get

-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]

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Therefore, the heat of formation of propane is 72.4 kJ/mol

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3 years ago
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