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kifflom [539]
2 years ago
15

The average yearly temperatures and latitudes of four cities are shown in the table above. Which of the following is demonstrate

d by the cities in the table?
a As latitude decreases, average yearly temperature generally increases.
B.
Climate is only influenced by latitude in the Southern Hemisphere.
C.
As latitude decreases, average yearly temperature generally decreases.
D.
The climate of an area is never influenced by the region's latitude.
Chemistry
2 answers:
Svet_ta [14]2 years ago
4 0

Answer:

.

C.

As latitude decreases, average yearly temperature generally

Tems11 [23]2 years ago
4 0

Answer:

Wanna

Explanation:

J. Oing a meet its going to be fun yzn-dggf-jcq

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2 years ago
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In which set do all elements tend to form anions in binary ionic compounds? li, na, k n, o, i c, s, pb k, fe, br?
pav-90 [236]

Atoms or molecule after gaining of electron possesses negative charge and is known as anion.

For the given sets:

  • Li, Na, K

The given elements are alkali metals and have tendency to lose electrons easily and form cations.

  • N, O, I

The given elements are non-metals and are electronegative. So, they gain electrons easily and form anion.

  • C, S, Pb

Carbon has tendency to form bond by sharing of electrons, Sulfur has tendency to gain electrons and form anion whereas Lead has tendency to lose electron.

  • K, Fe, Br

Potassium and Iron has tendency to lose electron and form cation whereas Bromine has tendency to gain electron to form anion.

Hence, from the given sets, all elements of set: N, O, I have tendency to form anions in binary ionic compounds.

8 0
3 years ago
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Burning coal produces which type of gas?
vovikov84 [41]

Answer:

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Explanation:

carbon dioxide is the primary greenhouse gas produced by burning fossil fuels like coal and oil

5 0
2 years ago
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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

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3 years ago
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