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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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Does N2F2 obey the octet rule? Explain.
zavuch27 [327]
It indeed does because 5(2) +7(2) =24
4 0
2 years ago
A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the
Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

Initial temperature of the sample = 95.0 °C

Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q =  m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C

-4784 * c(aluminium) = -4184

c(aluminium) = 0.875 J /g°C

The specific heat of aluminium is 0.875 J/g°C

7 0
3 years ago
What organisms can be infected by Ebola ​
Alenkasestr [34]
Human organisms, and animals can also be infected by Ebola
8 0
3 years ago
Read 2 more answers
HELP ME PLEASE!!!!
liq [111]

Answer:

1.D

2.A

3.C

4.A

Explanation:

6 0
3 years ago
Read 2 more answers
46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride
slega [8]

Answer:

m_{HgCl_2}=42.7gHgCl_2

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2

Best regards.

3 0
3 years ago
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