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3 years ago

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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What will the presence of H+ ions in a solution cause?

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H+ ions are proton charged ions that are present

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3 years ago

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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

Misha Larkins [42]

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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2 years ago

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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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Welding and computer chips

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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

q = 17.38 kJ

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

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

Add up all the energy gives 157.8 kJ

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3 years ago