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3 years ago

How does environmental changes affect animals?

Chemistry

1 answer:

Firdavs [7]3 years ago

3 0

Answer:

Humans and wild animals face new challenges for survival because of climate change. More frequent and intense drought, storms, heat waves, rising sea levels, melting glaciers and warming oceans can directly harm animals, destroy the places they live, and wreak havoc on people's livelihoods and communities.

Explanation:

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Which arrow represents the activation energy for the forward reaction

ValentinkaMS [17]

Answer:

Explanation:

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1 year ago

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Would you expect an atom to have a charge?

puteri [66]

Answer:

An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral.

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8 months ago

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2 years ago

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How many H2O molecules are in 183.2 grams of H20 gas?

jek_recluse [69]

Answer: There are $61.24 \times 10^{23}$ molecules present in 183.2 grams of $H_{2}O$ gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of $H_{2}O$ are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol$

According to the mole concept, there are $6.022 \times 10^{22}$ molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

$10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}$

Thus, we can conclude that there are $61.24 \times 10^{23}$ molecules present in 183.2 grams of $H_{2}O$ gas.

6 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago