Answer:
total number of electron in 1 litter is 3.34 × 
electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 ×
molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 × × 1
total number of electron in water is = 6.02×
electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 × 
m³
mass of 1 litter = 1000 × 1 ×
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter × 
total number of electron in 1 litter = 1000 × 
total number of electron in 1 litter is 3.34 × electron
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
<em></em>
Read more: brainly.com/question/8898885
Psychology on Egenuity Oct 5th 2018 says answer is C
Potential difference is the work done In moving a charge from one point to another in a conductor
Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:
where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,
<u>P = 2439.5 W = 2.439 KW</u>
