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11111nata11111 [884]
3 years ago
12

The dipole moment of hf is 1.91d. what is the dipole moment of hf in c⋅m? express your answer in coulomb-meters to three signifi

cant figures.
Chemistry
1 answer:
Drupady [299]3 years ago
3 0
<span>6.37x10^-30 coulomb-meters The unit for dipole moments is the Debye which is 1x10^-21 C*m^2/s divided by the speed of light (299792458 m/s). So the conversion factor is 1x10^-21 C*m^2/s / 299792458 m/s = 3.33564095E-30 C*m And converting 1.91 d to Cm gives 1.91 * 3.33564095E-30 C*m = 6.37x10^-30 coulomb-meters</span>
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borishaifa [10]

Answer:

1.20 M

Explanation:

Convert grams of Na₂CO₃ to moles.  (50.84 g)/(105.99 g/mol) = 0.4797 mol

Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M

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Which of the following substances (with specific heat capacity provided) would show the greatest temperature change upon absorbi
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Answer:

Pb is the substance that experiments the greatest temperature change.

Explanation:

The specific heat capacity refers to the amount of heat energy required to raise in 1 degree the temperature of 1 gram of substance. The highest the heat capacity, the more energy it would be required. These variables are related through the equation:

Q = c . m . ΔT

where,

Q is the amount of heat energy provided (J)

c is the specific heat capacity (J/g.°C)

m is the mass of the substance

ΔT is the change in temperature

Since the question is about the change in temperature, we can rearrange the equation like this:

\Delta T = \frac{Q}{c.m}

All the substances in the options have the same mass (m=10.0g) and absorb the same amount of heat (Q=100.0J), so the change in temperature depends only on the specific heat capacity. We can see in the last equation that they are inversely proportional; the lower c, the greater ΔT. Since we are looking for the greatest temperature change, It must be the one with the lowest c, namely, Pb with c = 0.128 J/g°C. This makes sense because Pb is a metal and therefore a good conductor of heat.

Its change in temperature is:

\Delta T = \frac{q}{c.m} = \frac{100.0 J}{0.128 J/g.C . 10.0g } = 78.1

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SOVA2 [1]
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A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope.
dexar [7]
The answer is 3.

<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:

</span>(1/2) ^{n} =x

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x = 0.125


Now, let's implement this in the equation: 

<span>(1/2) ^{n} =0.125
</span>
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Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>
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