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Dovator [93]
3 years ago
11

Consider 70.0-g samples of two different compounds consisting of only sulfur and oxygen. One of the compounds consists of 35.0 g

of sulfur, and the other has 28.0 g of sulfur. Determine the ratio in whole numbers of the masses of sulfur that combine with 1.00 g of oxygen between the two compounds.
Chemistry
1 answer:
Mariana [72]3 years ago
5 0

Answer:

1 : 1.5

Explanation:

First Sample;

Ratio of sulfur and Oxygen

Mass of sulfur : Mass of oxygen

Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 35 = 35g

35g : 35g

1 : 1

Second Sample;

Ratio of sulfur and Oxygen

Mass of sulfur : Mass of oxygen

Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 28 = 42g

28g : 42g

1 : 1.5

Further reducing it to make oxygen 1;

0.6667 : 1

ratio in whole numbers of the masses of sulfur that combine with 1.00 g of oxygen between the two compounds;

0.6667 : 1

1 : 1.5

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If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit
Blizzard [7]

<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

<u>Explanation:</u>

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

V=10.00\times 7.25\times 3.00=217.5cm^3

  • <u>For a:</u>

To convert the volume of cuboid into milliliters, we use the conversion factor:

1mL=1cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL

Hence, the volume of the box is 217.5 mL

  • <u>For b:</u>

To convert the volume of cuboid into liters, we use the conversion factor:

1L=1000cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L

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3 0
3 years ago
If you are calculating the grams to mL ratio, you are trying to find the object’s…
Alex

Answer:

Density

Explanation:

The ratio of mass to the volume of an object is called its density. Unit of mass is grams and that of volume is mL.

Density = mass/volume

\text{Density}=\dfrac{\text{grams}}{\text{mL}}\\\\=\text{grams/mL}

If you are calculating the grams to mL ratio, it means that we are trying to find the object's density.

4 0
3 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

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Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

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1 g = 0.001 kg

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7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

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