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2 years ago

11

The climate of an area can be different from its weather. Which of the following statements describes the climate of an area?

1 answer:

KatRina [158]2 years ago

5 0

Answer:

A

Explanation:

because on the first alphabet they've been specific of the weather partaking in the area

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Balance this chemical formula<br> Fe + Cl2 = FeCl3

san4es73 [151]

First you find out how much each element has

Fe=1 * 2 Fe=1
Cl=2 *3 Cl=3 *2

now we multiply each so we can balance each side.

So now we get our balanced equation
2 Fe + 3 Cl2 = 2 FeCl<span>3</span>

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Predict the energy trend as a substance changes from a solid to liquid to gas on this graph.

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Answer:
Option-A (<span> It would increase from bottom left to top right) is the correct answer.

Explanation:
As we know converting solids into liquids and converting liquids into gases require energy. This energy provided increases the energy of the state and its particles start moving with higher velocities. Therefore, the energy of solids will be lower than liquids and gases respectively. While, liquids have greater energy than solids but less energy than gases. And, gases are the most energetic than solids and liquids.</span>

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24. How many electrons can each p orbital hold?

Dennis_Churaev [7]

Answer: Each p orbital can hold 6 electrons.

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3 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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3 years ago

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