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Crazy boy [7]
3 years ago
11

When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 10.085 g at 20ºC. If the d

ensity of water is taken to be 0.9975 g/mL at 20ºC, what is the percent error for the 10 mL of water.
Chemistry
1 answer:
Arte-miy333 [17]3 years ago
8 0
The percentage error is the error of the measured value to the true value. To find he percent error, the equation is as follows:

Percent error = |Measured Value - True Value|/True Value  * 100
The || is needed to get the absolute value of the difference. Substituting the values,

Percent error = |(10.085 g/10 mL) - 0.9975 g/mL|/<span>0.9975 g/mL  * 100
</span><em>Percent error = 1.1% </em>
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Sulfur and fluorine form several different compounds including sulfur hexafluoride and sulfur tetrafluoride. Decomposition of a
svetoff [14.1K]

Answer:

See Explanation

Explanation:

For SF6;

Since;

1.25 g of S corresponds to 4.44g of F

1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55

For SF4;

Since;

1.88 g of S corresponds to 4.44g of F

1 g of sulphur corresponds to 1 * 4.44/ 1.88  = 2.36

Hence;

Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4

=

3.55/2.36 = 1.5

Hence the law of multiple proportion is obeyed here.

3 0
3 years ago
Please help me I need help with this questions I’m very confused fused as to what the answer is please
zepelin [54]
Answer: This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
3 0
3 years ago
Create the Equation: What is the Percent Yield of Ammonia (NH3) if 11.8 g is recovered in a reaction with 7.02 x 10^23 molecules
insens350 [35]

Answer:

Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100

%

yield.

The balanced chemical equation

N

2

(

g

)

+

3

H

2

(

g

)

→

2

NH

3

(

g

)

tells you that every

1

mole of nitrogen gas that takes part in the reaction will consume

3

moles of hydrogen gas and produce

1

mole of ammonia.

In your case, you know that

1

mole of nitrogen gas reacts with

1

mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

what you need

3 moles H (sub 2)

>

what you have

1 mole H (sub2)

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume

1

mole of hydrogen gas and produce

1

mole H

2

⋅

2 moles NH

3

3

moles H

2

=

0.667 moles NH

3

at

100

%

yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced

0.50

moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every

100

moles of ammonia that could theoretically be produced.

You know that

0.667

moles will produce

0.50

moles, so you can say that

100

moles NH

3

.

in theory

⋅

0.50 moles NH

3

.

actual

0.667

moles NH

3

.

in theory

=

75 moles NH

3

.

actual

Therefore, you can say that the reaction has a percent yield equal to

% yield = 75%

−−−−−−−−−−−−−

or 75 moles NH sub3

I'll leave the answer rounded to two sig figs.

5 0
3 years ago
1. The emission spectrum of mercury atoms has a bright green line with wavelength
SashulF [63]

Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.

From the formula;

E = hc/λ

h = Plank's constant =6.6 * 10^-34 Js

c = speed of light=  3 * 10^8

λ = wavelength = 546.1 * 10^-9 m

E =  6.6 * 10^-34 * 3 * 10^8/546.1 * 10^-9

E =3.63 * 10^-19 J

Also;

E =hf

Where;

h = Planks's constant

f = frequency of photon

f = E/h

f = 3.63 * 10^-19 J/6.6 * 10^-34

f = 5.5 * 10^14 Hz

Learn more: brainly.com/question/18415575

5 0
2 years ago
3. What do the most abundant elements in Earth's atmosphere have in common?​
Advocard [28]

Answer:

They all have: nitrogen :780,900 755,100

oxygen: 209,500 231,500

argon 9,300 12,800

carbon dioxide

386 591

8 0
3 years ago
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