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Afina-wow [57]
1 year ago
7

A student wants to conduct an experiment to find out how pulse rate changes as the length of time spent exercising increases the

dependent variable will be
Chemistry
1 answer:
tester [92]1 year ago
3 0

Answer:

I'm not 100% percent sure what your question is, but  here is what I think you are asking.

Explanation:

The dependent variable is pulse rate and the indepedent variable is the amount of exericse. The dependent variable changes as the independnet variable changes.

You might be interested in
6.
RUDIKE [14]

Answer:

THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.

Explanation:

Equation:

Al3+ + 3e- -------> Al

3 F of electricity is required to produce 1 mole of Al

3 F of electricity = 27 g of Al

If 18 g of aluminium was used, the quantity of electricity to be used up will be:

27 g of AL = 3 * 96500 C

18 G of Al = x C

x C = ( 3 * 96500 * 18 / 27)

x C = 193 000 C

For 18 g of Al to be produced, 193000 C of electricity is required.

To calculate the current required to produce 193 000 C quantity of electricity, we use:

Q = I t

Quantity of electricity = Current * time

193 00 = I * 1.50 * 60 * 60 seconds

I = 193 000 / 1.50 * 60 *60

I = 193 000 / 5400

I = 35.74 A

The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A

3 0
1 year ago
A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of
Mars2501 [29]

<u>Answer:</u> The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=0.3986g

Mass of H_2O=0.0578g

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, \frac{12}{44}\times 0.3986=0.1087g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, \frac{2}{18}\times 0.0578=0.0066g of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100      ......(1)

  • <u>For Carbon:</u>

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%

  • <u>For Hydrogen:</u>

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

8 0
1 year ago
A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it
patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0
1 year ago
Explain how your experimental data for Rf values are, or are not, consistent with your predictions of enantiomeric and diastereo
hichkok12 [17]

Answer:

Answer is explained below.

Explanation:

As (+) menthol and (-) menthol are enantiomers whose physical properties are same except optical activity so we can expect they have similar Rf values.

Whereas diastereomers have different physical properties and different Rf values.

For example when the (+) menthol , (-) menthol, isomenthol and neomenthol undergo TLC (thin layer chromatography) the

Rf values of.(+menthol) = .447

Rf (+isomenthol) = .395

Rf (+neomenthol)= .487

Rf (-menthol) = .434

The above data shows that (+) menthol and (-) menthol have almost same Rf values and vary a little i.e 0.447 and 0.437. So we can conclude them as enantiomers

Whereas (+) menthol or (+) neomenthol or (+) isomenthol i.e 0.447 , 0.395 and 0.487 have different Rf values. We can conclude them as diasteromers.

(+) menthol and (-) menthol - enantiomers

(+) menthol and (+) neomenthol- diastereomers

(-) menthol and (+) isomenthol - diastereomers

3 0
1 year ago
Read 2 more answers
A gas occupies a volume of 72 ml at 400 k and 800 torr. if the temperature drops to 200 k and the pressure changes to 400 torr,
satela [25.4K]
We are tasked to solve for the volume of the gas that occupies when pressure and temperature changes to 400 Torr and 200 Kelvin from Torr and 400 Kelvin. We can use ideal gas law assuming constant gas composition and close system. The solution is shown below:
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1P2
V2 = 800*72*200 / 400*400
V2 = 72 ml

The answer for the volume is 72 ml.
7 0
2 years ago
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