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3 years ago

What do you think would be the most challenging part of investigating a fire or explosion crime scene? Why?

Chemistry

1 answer:

Lena [83]3 years ago

7 0

Answer:

I think finding the source of the fire would be the most difficult aspect seeing as though the fire would have burned any evidence

Explanation:

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Starting with a vessel of temperature 50.1°C, a pressure of 6263.MmHg and volume 461.1mL, what is its final volume, in liters, i

il63 [147K]

Answer:

The value of final volume inside the vessel $V_{2}$ = 5.17 ml

Explanation:

Initial pressure $P_{1}$ = 6263 mm Hg = 8.24 atm = 835 K pa

Initial temperature $T_{1}$ = 50.1 ° c = 323.1 K

Initial volume $V_{1}$ = 461.1 ml = 0.0004611 $m^{3}$

Final temperature $T_{2}$ = - 95.8 ° c = 177.2 K

Finial pressure $P_{2}$ = 411 atm = 41644.6 K PA

We know that

$P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }$

Put all the values in the above equation

⇒ 835 × $\frac{0.0004611}{323.1}$ = 41644.6 × $\frac{V_{2} }{177.2}$

⇒ $V_{2}$ = 5.07 × $10^{-6}$ $m^{3}$

⇒ $V_{2}$ = 5.17 ml

This is the value of final volume inside the vessel.

8 0

3 years ago

Convert: 20.9 mL = ___L 2.09 0.209 0.0209 209 20.9

aleksklad [387]

Answer:

0.209

Explanation:

when dealing with mL to L the rule of thumb seems to be to move your decimal place back two spaces.

6 0

3 years ago

A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

5 0

2 years ago

The solubility of Z is 60 g/ 100 g water at 20 °C. How many grams of solution are produced when a saturated solution is prepared

V125BC [204]

Answer:

Saturated solution = 180 gram

Explanation:

Given:

Solubility of Z = 60 g / 100 g water

Given temperature = 20°C

Amount of water = 300 grams

Find:

Saturated solution

Computation:

Saturated solution = [Solubility of Z] × Amount of water

Saturated solution = [60 g / 100 g] × 300 grams

Saturated solution = [0.6] × 300 grams

Saturated solution = 180 gram

3 0

3 years ago

When assigning electrons to orbitals, which would be the most likely 'address' for the next electron following the 5s2 electron?

Bess [88]

The answer is number 3, 4d1.

4 0

4 years ago

Read 2 more answers