Which of the following is the coldest climate zone?

Chemistry

2 answers:

Phantasy [73]3 years ago

7 0

Answer:

Polar zone

I hope it helps

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#Captainpower

pls mark me brainliest

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dolphi86 [110]3 years ago

6 0

Answer: polar
explanation:

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Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.

zavuch27 [327]

15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.

Step 1. Convert grams of O_2 to moles of O_2

Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2

Step 2. Use the molar ratio of HgO:O_2 to convert moles of O_2 to moles of HgO

Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = 15.63 mol HgO

7 0

4 years ago

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LenKa [72]

In a non-flowering plant, the embryo is in spores found in the stem, and in a flowering plant, the embryo is in seeds found in the flower.
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6 0

3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$