Answer:
Mass of oxygen in glucose = 29.3g
Explanation:
Mass of glucose given is 55grams.
We are to find the mass of oxygen in this compound.
In the compound we have 6 atoms of oxygen.
Solution
To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.
Atomic mass of C = 12g
H = 1g
O = 16g
Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180
Mass of O in glucose = 
x 55
= 
x 55
= 0.53 x 55
Mass of oxygen in glucose = 29.3g
Waves interact with matter in several ways. The interactions occur when waves pass from one medium to another. Besides bouncing back like an echo, waves may bend or spread out when they strike a new medium. These three ways that waves may interact with matter are called reflection, refraction, and diffraction.
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
Answer:
It would take 3.11 J to warm 3.11 grams of gold
Explanation:
Step 1: Data given
Mass of gold = 3.11 grams
Temperature rise = 7.7 °C
Specific heat capacity of gold = 0.130 J/g°C
Step 2: Calculate the amount of energy
Q = m*c*ΔT
⇒ Q = the energy required (in Joules) = TO BE DETERMINED
⇒ m = the mass of gold = 3.11 grams
⇒ c = the specific heat of gold = 0.130 J/g°C
⇒ ΔT = The temperature rise = 7.7 °C
Q = 3.11 g * 0.130 J/g°C * 7.7 °C
Q = 3.11 J
It would take 3.11 J to warm 3.11 grams of gold
