Answer : The volume for 6.0m HCl solution required = 62.5 ml
Solution : Given,
Initial concentration of HCl solution = 6.0m
Final concentration of HCl solution = 1.5m
Final volume of HCl solution = 250 ml
Initial volume of HCl solution = ?
Formula used for dilution is,
where,
= initial concentration
= final concentration
= initial volume
= final volume
Now put all the given values in above formula, we get the initial volume of HCl solution.
= 62.5 ml
Therefore, the volume for 6.0m HCl solution required = 62.5 ml
x = 20 long tables
y = 5 round table
Explanation:
We have the following system of equations:
x + y = 25
8x + 6y = 190
From the first equation we have:
x = 25 - y
And we replace x in the second equation:
8(25 - y) + 6y = 190
200 - 8y + 6y = 190
200 - 2y = 190
200 - 190 = 2y
10 = 2y
y = 5
Now we insert the value of y in the next equation:
x = 25 - y
x = 25 - 5
x = 20
Learn more about:
system of equations
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Answer:
20.0 grams
Explanation:
If the density of gold is 20.0 g/mL, then we can multiply it by 1 mLto find the weight of 1 mL of gold.
20.0
*1mL=20.0 grams
The balanced chemical reaction is written as:
<span>4C(s) + S8(s) → 4CS2(l)
We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.
</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8
The limiting reactant would be S8. We use this amount to calculate.
0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2
