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3 years ago

Warm and cold fronts are formed by the movement of different air masses.

Chemistry

1 answer:

vekshin13 years ago

3 0

That is true because the warm air goes up and the cold air comes down

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Taking a test please help. Thank you.

Komok [63]

It’s the 3rd one obviously bro

3 0

3 years ago

An electrochemical cell has the following standard cell notation.

Alona [7]

Explanation:

The given following standard cell notation.

Mg(s) | Mg^2+ (aq) || Aq^+(aq) | Aq(s)

Oxidation:

$Mg(s)\rightarrow Mg^{2+}+2e^-$ ....(1)

Magnesium metal by loosing 2 electrons is getting converted into magnesium cation. Hence, getting oxidized

Reduction:

$Ag^+(aq)+1e^-\rightarrow Ag(s)$ ...(2)

Silver ion by gaining 1 electrons is getting converted into silver metal. Hence, getting reduced.

Overall redox reaction: (1)+2 × (2)

$Mg(s)+2Ag^+(aq)\rightarrow Mg^{2+}+2Ag(s)$

4 0

3 years ago

Help please!!!!!!!!!

Hatshy [7]

Answer:

D. Element

Explanation:

Fe is an element. It isn't more than one element chemically combined (compound), it isn't a charged particle (ion), and it isn't the smallest possible unit of multiple atoms combined together (molecule).

3 0

3 years ago

When turgor pressure is high enough in a cell, the cell walls become _____

iren2701 [21]

Answer:

the answer is "Firm".

Explanation:

6 0

3 years ago

Read 2 more answers

If 62.3 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.862 g of precipitate, wh

Alexxx [7]

Answer:

The molarity of the lead(II) ion in the original solution is 0.03M

Explanation:

Step 1: The balanced reaction

Pb(NO3)2(aq) + 2 NaI aq) → PbI2(s) + 2 NaNO3 (aq)

Pb2+ + 2I- →PbI2

This means that for 1 mole Pb2+ consumed, there is needed 2 moles of I- to produce 1 mole of PbI2

Step 2: Calculate moles of PbI2

Moles of PbI2 = 0.862g / 461.01 g/mole

moles of PbI2 = 0.00187 moles

Step 3: calculate moles of Pb2+

For 1 mole of PbI2 produced we need 1 mole of Pb2+

This means for 0.00187 moles of PbI2 we need 0.00187 moles of Pb2+

Step 4: Calculate the molarity of the Pb2+ ion

Molarity of Pb2+ = moles of Pb2+ / volume =

Molarity of Pb2+ = 0.00187 moles / 62.3*10^-3

Molarity of Pb2+ ion = 0.03M

The molarity of the lead(II) ion in the original solution is 0.03M

5 0

3 years ago