Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
Answer:
a.3-5 is the modal class.
2 O2 + CH4 CO2 + 2 H2O
<span>What mass of CH4 is required to completely react with 100 grams of O2? </span>
<span>mass CH4 = 100 g O2 x (1 mol O2 / 32 g) x (1 mol CH4 / 2 mol O2) x (16.05 g / 1 mol CH4) </span>
<span> 25 grams CH4
</span>
Answer: D. The energy required to boil a substance
Explanation: I just took the quiz and got it correct :)
Answer:
cant understand the grammer rewrite and ill answer
Explanation:
