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hichkok12 [17]
3 years ago
12

If the same index fossils are found in different rock strate miles apart, what is probably true about the rock layers?

Chemistry
1 answer:
vodka [1.7K]3 years ago
6 0

Answer:

they will be different

Explanation:

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H+ ions are proton charged ions that are present
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The coolant in automobiles is often a 50/50 % by volume mixture of ethylene glycol, C2H6O2, and water. At 20°C, the density of e
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Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = \frac{mass}{molar mass}

                                                  = \frac{density \times volume}{molar mass}

                         = \frac{1.1088 \times 50}{62 g/mol}

                         = 0.894 mol

Hence, number of moles of water = \frac{50 \times 0.998}{18}

                                             = 2.77

As glycol is dissolved in water.

So, the molality = 0.894 \times \frac{1000}{49.92}

                           = 17.9

Therefore, the expected freezing point = -1.86 \times 17.9

                                                                  = -33.31^{o}C

Thus, we can conclude that the expected freezing point is -33.31^{o}C.

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3 years ago
Iodine-138 decays by beta decay. What element does it produce?
iVinArrow [24]

{xe}^{138}

Explanation:

the bottom number would be 54. This is because you add one to 53 to make it 54 and you add 0 to 138.

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Which of the following do you think use plasma? welding rocket engines computer chips surgery chemical analysis ozone generation
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How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Spe
dexar [7]

Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

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And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

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Add up all the energy gives 157.8 kJ

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