Answer:
12.19 hours prior to discovery was the corpse placed in the desert.
Explanation:
Heat required for fly eggs to develop into first instar larvae = 
From the given,
Exposure time = 16 hrs
Temperature = 27 C
The day time exposed hours is 10 hrs.-(7.00am-5.00pm)
So, the extra heat = 427.2-378= 49.2
The addition heat divided by average temperature.
Average temperature = 22.4 C
So, the total time exposed is night time hours to day time hours.
= 10+2.19 = 12.19 hrs.
Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.
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
Hello!
The H₃O⁺ concentration can be found using the definition of pH and clearing the equation for [H₃O⁺]. The solution has a pH lower than 7, so the Sauvignon Blanc is
acid. The calculation for [H₃O⁺] is shown below:
![pH=-log [H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%20%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}]= 10^{-pH}=10^{-3,24}=0,00058M](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%5D%3D%2010%5E%7B-pH%7D%3D10%5E%7B-3%2C24%7D%3D0%2C00058M%20)
So, the concentration of H₃O⁺ in a Sauvignon Blanc with a pH of 3,24 is
0,00058 MHave a nice day!
Answer:
a)23.2 L
b)68.3kPa
c)7.5 atm
d)60.5L
e)1.67 atm
Explanation:
From Boyle's law:
P1V1=P2V2
P1= 748mmHg
P2=725mmHg
V1= 22.5L
V2??
V2= P1V1/P2= 748×22.5/725= 23.2 L
b)
V1=4.0L
P1= 205×10^3Pa
V2= 12.0L
P2=???
P2= P1V1/V2= 205×10^3×4/12
P2= 68.3×10^3 Pa or 68.3kPa
c)
P1= 1 atm
V1= 196.0L
P2= ??
V2= 26.0L
P2= P1V1/V2=1×196.0/26.0
P2= 7.5 atm
d)
V1= 40.0L
P1= 12.7×10^3Pa
V2=???
P2= 8.4×103Pa
V2= P1V1/P2= 12.7×10^3×40.0/8.4×103
V2=60.5L
e)
V1= 100mL
P1= 1atm
V2= 60mL
P2=???
P2= P1V1/V2= 1×100/60
P2= 1.67 atm
I think the awnser to your question is C
