Answer:
1.200g
Explanation:
At the top it's 0 and that scale goes by 10s
The middle scale is 1 and it goes by 1s
The bottom scale is .2 and it goes by .1s
1+.2= 1.2 the extra zeroes are just place holders
1.200g
<span>6.12<span>(<span>1024</span>)</span></span><span>=<span><span>(6.12)</span><span>(<span><span>1e</span>+24</span>)</span></span></span><span>=<span><span>6.12e</span>+24</span></span>
<span>
=
</span>
Answer:
Balanced forces are forces that act in opposite directions and cancel each other out
let me know if you have any other questions
Answer:
3853 g
Step-by-step explanation:
M_r: 107.87
16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹
1. Calculate the moles of Ag₂S
Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S
2. Calculate the moles of Ag
Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag
3. Calculate the mass of Ag
Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag
You must react 3853 g of Ag to produce 567.9 kJ of heat
Answer:

Explanation:
In this case, we have to start with the <u>chemical reaction</u>:

So, if we start with <u>10 mol of cyclohexanol</u> (
) we will obtain 10 mol of cyclohexanol (
). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

With this value we can calculate the grams:

Now, we have as a product 500 mL of
. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

Finally, with these values we can calculate the <u>yield</u>:
%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!