Answer:
25.907°C
Explanation:
In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C
The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.
Thus:
= heat of combustion of benzoic acid
= heat energy released to water
= heat energy released to the calorimeter
Therefore,
![-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})](https://tex.z-dn.net/?f=-m_%7Bcombust%7D%2AH_%7Bcombust%7D%20%3D%20%5Bm_%7Bwater%7D%2Ac_%7Bwater%7D%20%2B%20C_%7Bcalori%7D%5D%2A%28T_%7Bf%7D%20-%20T_%7Bi%7D%29)
1.056*26.42 = [0.987*4.18 + 6.66](
- 23.32)
27.8995 = [4.12566+6.660]( - 23.32)
( - 23.32) = 27.8995/10.7857 = 2.587
= 23.32 + 2.587 = 25.907°C
In the modern periodic table the elements are arranged in order of increasing atomic number.
Answer:
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Explanation:
Using the formula
Ca Va = Cb Vb
Cb = 0.32 M
Vb = 50 mL = 50/1000 = 0.050L
Ca = 0.5 M
Va =?
Substituting for Va in the equation, we obtain:
Va = Cb Vb / Ca
Va = 0.32 * 0.05 / 0.5
Va = 0.016 / 0.5
Va = 0.032 L
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Viruses are not classified as being alive in because of the fact that they can't reproduce.
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
