Answer:
Hi ,
Answer:
A chemical energy is changed to a electrical energy
Answer:
The answer to your question is
1.- Volume = 3.4 ml
2.- Volume = 0.61 ml
3.- Mass = 2872.8 pounds
Explanation:
Problem 1
Volume = 18 ml
mass = 35.6 g
density = 10.5 g/ml
Process
1.- Calculate the volume of silver
Formula

solve for volume

Substitution

<u>volume = 3.4 ml</u>
2.- Problem 2
Total volume = ?
Volume = 18 + 3.4
Volume = 21.4 ml
Data
mass = 8.3 g
density = 13.6 g(ml
volume = ?
Formula

Solve for volume

Substitution

Result
<u>volume = 0.61 ml</u>
3.- Problem 3
Data
volume = 345 gal
density = 1 g/ml
mass = ?
Formula

Solve for mass
mass = density x volume
Covert gal to ml
1 gal --------------- 3785 ml
345 gal ------------- x
x = (345 x 3785) / 1
x = 1305825 ml
Substitution
mass = 1 x 1305825
mass = 1305825 g
Convert g to pounds
1 g ------------------- 0.0022 pounds
1305825 g ---------------- x
x = (1305825 x 0.0022)
<u> x = 2872.8 pounds</u>
4.00 moles of sodium has a mass of 91.96 grams.
To find an element's molar mass, you can use its atomic weight.
Answer:
True
Explanation:
Neutral solutions maintain a pH of 7. Water and human blood are great examples of neutral solutions. Acids mixed with bases can be neutralized and given a pH of 7.
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