Answer:
$1246.90
Explanation:
Since the bike lost 
% of it's value and it now currently at $
, we have to do 20% * $1039 to find the amount of money lost. 20%*1039=207.8. We have to add it up to find the original value so 1039+207.8=$1246.8
Im not really sure what your asking.... <span>Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, </span>14.70 pounds per square inch<span>, 1,013.25 × 10 </span>3<span> dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.
</span><span>""atmospheric pressure | Britannica.com""</span>
Answer:
Percent error = 1.5%
Explanation:
Given data:
Measured value of density of graphite = 2.3 g/cm³
Percent error = ?
Solution:
Formula:
Percent error = [Measured value - Actual value / actual value] × 100
Actual/accepted value of density of graphite = 2.266 g/cm³
Now we will put the values:
Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100
Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100
Percent error = 0.015 × 100
Percent error = 1.5%
The molecular mass of Carvone is calculated as;
= 12 (C)₁₀ + 1.008 (H)₁₄ + 16 (O)
= 120 + 14.112 + 16
= 150.112
%age of Carbon;
= (120 ÷ 150.112) × 100
= 79.94 %
%age of Hydrogen;
= (14.112 ÷ 150.112) × 100
= 9.40 %
%age of Oxygen;
= (16 ÷ 150.112) × 100
= 10.65 %
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
