Answer:
yes it is true.
Ice floats on the surface of water because ice is less dense than liquid water
Explanation:
Answer:
869 g Cl₂O
Explanation:
To find the theoretical yield of Cl₂O, you need to (1) convert moles SO₂ to moles Cl₂O (via mole-to-mole ratio from reaction coefficients) and then (2) convert moles Cl₂O to grams Cl₂O (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs of the given amount (10.0 moles).
1 SO₂ (g) + 2 Cl₂ (g) ----> 1 SOCl₂ (g) + 1 Cl₂O (g)
Molar Mass (Cl₂O): 2(35.453 g/mol) + 15.998 g/mol
Molar Mass (Cl₂O): 86.904 g/mol
10.0 moles SO₂ 1 mole Cl₂O 86.904 g
------------------------ x ---------------------- x ------------------ = 869 g Cl₂O
1 mole SO₂ 1 mole
It takes 21.3 days
<h3>Further explanation</h3>
Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
moles of glucose to produce 108.7 moles ethanol :
54.35 moles = 5 hours
moles of 1000 kg of glucose :
So for 5555.5 moles, it takes :
Answer:
Answer below
Explanation:
Just draw a photo of someone pushing an object across a table. Your push is the force acting on the object you're pushing.
Answer:
0.78 M
Explanation:
First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:
N2(g) + O2(g) <------> 2NO(g) Kc = ?
Writting Kc:
Kc = [NO]² / [N2] * [O2]
Replacing the given values we have then:
Kc = (0.6)² / (0.2)*(0.2)
Kc = 9
Now that we have the Kc, let's see what happens next.
We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:
N2(g) + O2(g) <------> 2NO(g)
I: 0.2 0.2 0.9
C: +x +x -2x
E: 0.2+x 0.2+x 0.9-2x
Replacing in the Kc expression we have:
Kc = [NO]² / [N2] * [O2]
9 = (0.9-2x)² / (0.2+x)*(0.2+x) ----> (this can be expressed as 0.2+x)²
Here, we solve for x:
9 = (0.9-2x)² / (0.2+x)²
√9 = (0.9-2x) / (0.2+x)
3(0.2+x) = 0.9-2x
0.6 + 3x = 0.9 - 2x
3x + 2x = 0.9 - 0.6
5x = 0.3
x = 0.06 M
This means that the final concentration of NO will be:
[NO] = 0.9 - (2*0.06)
[NO] = 0.78 M
