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Luba_88 [7]
3 years ago
10

Ling decided to do a science project on exercise. She wanted to find out how long it takes for a person's heart rate to slow dow

n to normal after exercise.Which of the following was likely the scientific prediction Ling made about her project?
A.
My heart beats 60 times each minute.
B.
Exercise is important.
C.
A person's heart rate slows down in three minutes.
D.
I found out that my heart beat slows down after five minutes.
Chemistry
1 answer:
forsale [732]3 years ago
8 0
C- a persons heart rate slows down after 5 minutes.
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What piece of equipment is used to measure: length
avanturin [10]

Answer:

Ruler

Explanation:

Ruler and eraser kakgjwjeigidiifigig

8 0
2 years ago
Read 2 more answers
Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol
pishuonlain [190]

<u>Answer:</u> The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

<u>Explanation:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g

  • <u>Calculating the freezing point:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.99°C/m

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62 g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC

Hence, the freezing point of solution is -117.54°C

  • <u>Calculating the boiling point:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_b = molal boiling point elevation constant = 1.20°C/m.g

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62  g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC

Hence, the boiling point of solution is 80.48°C

3 0
3 years ago
A. 20 moles of NH3 are needed to produce how many moles of H2O?
tiny-mole [99]

a. 30 moles of H₂O

b. 2.33 moles of N₂

<h3>Further explanation</h3>

Given

a. 20 moles of NH₃

b. 3.5 moles of O₂

Required

a. moles of H₂O

b. moles of N₂

Solution

Reaction

4NH₃+3O₂⇒2N₂+6H₂O

a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :

=6/4 x mol NH₃

= 6/4 x 20 moles

= 30 moles

b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :

=2/3 x mol O₂

= 2/3 x 3.5 moles

= 2.33 moles

7 0
2 years ago
determine the maximum amount of NaN03 that was produced during the experiment. Explain how you determined the amount
Angelina_Jolie [31]

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl  is the limiting reactant and 9 moles of NaNO3 is obtained.

7 0
3 years ago
What the first questions answer? Anyone know if so plz help a siesta out
Marysya12 [62]

Answer:

hi

Explanation:

8 0
3 years ago
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