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LuckyWell [14K]
2 years ago
7

There is water on the pan of the scale as you measure the mass of an object. If you were to ignore the water, what would be the

effect on your density calculation?
Chemistry
1 answer:
mihalych1998 [28]2 years ago
6 0
Remember that density refers to the "mass per unit volume" of an object.

So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.

In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is

As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:

density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml

The effect on density would be that it would erroneously appear to be greater

Hope this helps!

Good luck
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Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
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Vm - molar volume at STP.
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1 year ago
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What is the concentration of Iodine I2 molecules in a solution containing 2.54 g of iodine 250 cm3 of solution? A 0.01mol/dm3 B
gulaghasi [49]

Answer:

C. 0.04 moles per cubic decimeter.

Explanation:

The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine (c), measured in moles per cubic decimeter, can be determined by the following formula:

c = \frac{m}{M\cdot V} (1)

Where:

m - Mass of iodine, measured in grams.

M - Molar mass of iodine, measured in grams per mol.

V - Volume of solution, measured in cubic decimeters.

If we know that m = 2.54\,g, M = 253.809\,\frac{g}{mol} and V = 0.25\,dm^{3}, then the concentration of iodine in a solution is:

c = \frac{2.54\,g}{\left(253.809\,\frac{g}{mol} \right)\cdot (0.25\,dm^{3})}

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Hence, the correct answer is C.

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1 year ago
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

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Explanation:

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From the data give we can see that

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=>        E_{cell}__{AC}}=1.56- 0.62

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