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Pavlova-9 [17]
3 years ago
12

A cheetah can go from a state of resting to running at 20 m/s in just two seconds. What is the cheetah's average acceleration?

Chemistry
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

D is answer ..... !!!!!!!!.......!!!!!!#.#.....

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The name given to the Earth's climate when there was no ice at the poles
Citrus2011 [14]

Answer:

i think snowball, it sounds weird but its true (i think im sorry if its wrong)

Explanation:

7 0
3 years ago
If the internal energy of a system increases but there is no change in temperature, then the system's energy is increasing.
777dan777 [17]

Answer:

kintic

Explanation:

4 0
3 years ago
Read 2 more answers
Transferring or sharing electrons between atoms forms
Brums [2.3K]
Transferring or sharing electrons between atoms forms a covalent bond.<span> Covalent bonding is when atoms share electrons. It is a chemical bond that involves the sharing of electron pairs. These pairs are called bonding pairs. Examples of compounds that has covalent bonds are CO2, organic compounds, lipids and proteins.</span>

6 0
3 years ago
. In a titration, a 25.0 mL sample of 0.150 M HCl is neutralized with 44.45 mL of Ba(OH)2. a. Write the balanced molecular equat
choli [55]

Answer:

Equation of reaction:

a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O

b) Molarity of base = 0.042 M.

Explanation:

Using titration equation

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2

NB is the number of mole of base = 1

CA is the molarity of acid =0.15M

CB is the molarity of base = to be calculated

VA is the volume of acid = 25 ml

VB is the volume of base = 44.45mL

Substituting

0.15×25/CB×44.45 = 2/1

Therefore CB =0.15×25×1/44.45×2

CB = 0.042 M.

5 0
3 years ago
Please help quick
Rama09 [41]

Answer:

c = 0.898 J/g.°C

Explanation:

1) Given data:

Mass of water = 23.0 g

Initial temperature = 25.4°C

Final temperature = 42.8° C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.18 J/g°C

ΔT = 42.8°C - 25.4°C

ΔT = 17.4°C

Q = 23.0 g ×  × 4.18 J/g°C × 17.4°C

Q = 1672.84 j

2) Given data:

Mass of metal = 120.7 g

Initial temperature = 90.5°C

Final temperature = 25.7 ° C

Heat released = 7020 J

Specific heat capacity of metal = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25.7°C - 90.5°C

ΔT = -64.8°C

7020 J = 120.7 g ×  c ×  -64.8°C

7020 J = -7821.36 g.°C ×  c

c = 7020 J / -7821.36 g.°C

c = 0.898 J/g.°C

Negative sign shows heat is released.

7 0
3 years ago
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