HELP!!! PLEASE!!!

What electives should i take in highschool to prepare to take criminology?

dybincka [34]

Answer:

depending on what school you go to and what classes you are allowed to take in which grades, you should take forensics.

4 0

2 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

Best regards.

7 0

3 years ago

A cracker crushed in water will test positive with iodine but negative with benedict’s solution. A

german

A cracker which contains starch,is test positive with iodine solution but not Benedict's solution. This is because Benedict's solution is used to test for reducing sugars like glucose,galactose,fructose,maltose and lactose.In this case, the cracker is added with amylase enzyme which hydolyses starch into maltose,thus benedict's solution is test positive

7 0

3 years ago

If 4.65 LL of CO2CO2 gas at 22 ∘C∘C at 793 mmHg mmHg is used, what is the final volume, in liters, of the gas at 35 ∘C∘C and a p

otez555 [7]

Answer:

About 7.9 L.

Explanation:

We can utilize the ideal gas law. Recall that:

$\displaystyle PV = nRT$

Because the amount of carbon dioxide does not change, we can rearrange to formula to:
$\displaystyle \frac{PV}{T}= nR$

Because the right-hand side stays constant, we have that:
$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = nR$

Hence substitute initial values and known final values:
$\displaystyle \begin{aligned} \frac{(793\text{ mm Hg})(4.65 \text{ L})}{(22 \text{ $^\circ$C})} & = \frac{(743 \text{ mm Hg})V_2}{(35\text{ $^\circ$C})} \\ \\ V_2 & = 7.9\text{ L}\end{aligned}$

Therefore, the final volume is about 7.9 L.

8 0

2 years ago

You are a NASCAR pit crew member. Your employer is leading the race with 20 laps to go. He just finished a pit stop and has 5.0

jek_recluse [69]

Answer:

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

Explanation:

Density = mass / volume

Density of fuel = 700 g/ 1 gal

Therefore, the mass of fuel in 1 gallon = 700 g

The driver has 5.0 gallons of fuel in the tank.

The mass of 5.0 gallons of fuel = 5 × 700 = 3500 g of fuel

Equation of the combustion of fuel, C₅H₁₂ is given below:

C₅H₁₂ + 8 O₂ ---> 6 H₂O + 5 CO₂

1 mole C₅H₁₂ requires 8 moles of O₂

1 mole of C₅H₁₂ has a mass = 72 g

8 moles of O₂ has a mass = 256 g

Therefore, 300 g of O₂ will require 300 × (72/256) g of C₅H₁₂ = 84.375 g of C₅H₁₂

84.375 g of fuel is used by the car per lap;

20 laps will require 20 × 84.375 g of fuel = 1687.5 g of fuel.

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

7 0

3 years ago