Answer:
Number of moles = 2.89 mol
Explanation:
Given data:
Number of moles of sugar = ?
Mass of sugar = 990 g
Solution:
Formula:
Number of moles = mass/molar mass
Molar mass of C₁₂H₂₂O₁₁:
12× 12 + 22×1.008 + 16×11 = 342.2 g/mol
Number of moles = 990 g / 342.2 g/mol
Number of moles = 2.89 mol
The answer is; C
The radioisotopes in the interior of the earth are not in pure form and have many embedded impurities and are interspersed. Therefore when a neutron is released by a decaying atom, its chances of hitting another radioisotope atom (to continue the chain reaction) are lower than in a manmade nuclear reactor. This way, the radioactive chain reaction in the earth’s interior is not well sustained.
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be
200mL × 1g = 1000 mL × x(g)
x(g) =
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴
y(g) =
y(g) = 5g of benzalkonium chloride.
Now, at 17% concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
=
z(mL) =
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
Answer:
<h2>Density = 0.46 g/mL</h2>
Explanation:
Density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass = 5.52 g
volume = 12 mL
Substitute the values into the above formula and solve for the Density
That's
<h3>
</h3>
We have the final answer as
<h3>Density = 0.46 g/mL</h3>
Hope this helps you