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3 years ago

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3. How does temperature affect a football players performance?

1 answer:

castortr0y [4]3 years ago

7 0

Hello!

3. How does temperature affect a football players performance?

Independent Variable: Temperature

<em>Temperature is the independent variable because that is what you are testing.</em>

Dependent Variable: Points scored in game

<em>Since you are attempting to measure how well football players do in different climates, you must measure their scores.</em>

Constants: Uniform, field played in, and ball used

<em>If you use a different uniform, field, or ball, then the data is obviously going to be inconsistent.</em>

Hypothesis: If football players are in colder climates, then they perform better during games.

<em>This is quite self-explanatory: just use the latter data points to craft your hypothesis. Keep in mind that a hypothesis should NEVER be a question!</em>

<em />

I hope I helped! :-)

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Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated wi

julia-pushkina [17]

<u>Answer:</u> The $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL$

Putting values in above equation, we get:

$1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M$

The concentration of $Ca(OH)_2$ comes out to be 0.011 M.

The balanced equilibrium reaction for the ionization of calcium hydroxide follows:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

The expression for solubility constant for this reaction follows:

$K_{sp}=[Ca^{2+}][OH^-]^2$

Putting the values in above equation, we get:

$K_{sp}=(0.011)\times (2\times 0.11)^2$

$K_{sp}=5.324\times 10^{-6}$

Hence, the $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

6 0

3 years ago

Read 2 more answers

Compound A has 4 moles of hydrogen, 4 moles of oxygen, and

Sophie [7]

Answer:

H₄SiO₄

Explanation:

From the question given above, the following data were obtained:

Hydrogen (H) = 4 moles

Oxygen (O) = 4 moles

Silicon (Si) = 1 mole

Empirical formula =?

Since the number of mole of each atom of the element present is known, the empirical formula of the compound can be written as follow:

H₄SiO₄

3 0

3 years ago

A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v

miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution $C_1=?$

Volume of the sport drink before dilution $V_1=5.4 mL$

Concentration of red dye in sport drink after dilution $C_2=18.1 ppm$

Volume of the sport drink after dilution $V_2=25.0 mL$

$C_1V_1=C_2V_2$ ( dilution )

$C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}$

$C_1=83.80 ppm\approx 84.0ppm$

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

6 0

3 years ago

I need help fast!! Djdndndndndndnnds

adell [148]

Answer:

1st one: carbon dioxide

2nd: Glucose

3rd: Oxygen

4th: Energy

Sorry if this is wrong because I don't see the options!

Hope this helps!

6 0

3 years ago

Three moles of oxygen gas is stored in a 2.5-liter container at 35°C. What is the pressure of this gas?

nadezda [96]

Answer:

The pressure of the gas is 30.3072 atm.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= ?
V= 2.5 L
n= 3 moles
R= 0.082 $\frac{atm*L}{mol*K}$
T= 35 C= 308 K (being 0 C= 273 K)

Replacing:

P* 2.5 L= 3 moles* 0.082 $\frac{atm*L}{mol*K}$ *308 K

Solving:

$P=\frac{ 3 moles* 0.082 \frac{atm*L}{mol*K} *308 K}{2.5 L}$

P= 30.3072 atm

<u><em>The pressure of the gas is 30.3072 atm.</em></u>

4 0

3 years ago