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3 years ago

Which technique reveals latent fingerprints on wooden furniture?

Chemistry

2 answers:

love history [14]3 years ago

7 0

Answer: Nonporous

Explanation: I don’t know if this is right but please correct me!

balu736 [363]3 years ago

3 0

Answer: powder

Explanation: nonporus is the same or objects that require the powder technique to be used

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Ammonium nitrate is an ingredient in cold packs used by sports trainers for injured athletes. Calculate the change in temperatur

Ronch [10]

Answer:

See explanation

Explanation:

First, we need to use the correct expression:

Q = m*Cp*ΔT (1)

Q = n*ΔH (2)

These are the 2 expressions to calculate heat or energy.

Now, we want to know the change of temperature of the nitrate in water after being added, so with the innitial data of nitrate, we can calculate heat using the second expression. First, we need to calculate moles with the molecular mass:

n = m/MM

n = 42/80.1 = 0.52 moles

With these moles, we can calculate heat with the ΔH of this reaction:

Q = 0.52 * 25.7 = 13.364 kJ or 13,364 J

However, this heat as is being absorbed, the value would be negative.

Now that we have heat, we can use expression (1) and plug these values to solve for ΔT, but before, we need to know the total mass of the solution (water + nitrate)

m = 250 + 42 = 292 g

now, solving for ΔT:

-13,364 = 292 * 4.18 * ΔT

ΔT = -13,363 / (292 * 4.18)

ΔT = -10.95 °C

3 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$