Answer: 14.3%Explanation: In order to find the mass percent of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in 100 g of compound.
In your case, you know that an unknown mass of hydrogen reacts with 0.771 g of carbon to form 0.90 g of hydrocarbon, which is a compound that contains only carbon and hydrogen.
Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon.
Now, if 0.90 g of this compound contain 0.129 g of hydrogen, it follows that 100 g of this compound will contain.
So, if 100 g of this compound contain 14.33 g of hydrogen, it follows that the mass percent of hydrogen is 14.3%
<span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span><span><span>
</span></span>
Answer:
46.3g H2O
Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)
then use factor label method to solve
82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
Answer:
31395 J
Explanation:
Given data:
mass of water = 150 g
Initial temperature = 25 °C
Final temperature = 75 °C
Energy absorbed = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 75 °C - 25 °C
ΔT = 50 °C
now we will put the values in formula
q = m . c . ΔT
q = 150 g × 4.186 J/g.°C × 50 °C
q = 31395 J
so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .
For this problem we use the wave equation. It is expressed as the speed (c) is equal to the product of frequency (f) and wavelength (v).
c = v x f
We know the wavelength of the an red light which is 6.5 x 10^-7 m. Now, we solve for the wavelength of the unknown wave to see the relation between the two waves.
2.998 X 10^8 = 5.3 X 10^15 X v
v = 2.998 X 10^8 / (5.3 X 10^15) = 5.657 X 10^-8 m
Therefore, the wavelength of the unknown wave is less than the wavelength of the red light.
Mercury is the smallest and closest plant to the sun