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joja [24]
3 years ago
9

A miner develops cancer of the esophagus. ten years later

Chemistry
2 answers:
Free_Kalibri [48]3 years ago
4 0

Answer:

a miner or minor like young?

Explanation:

Dmitriy789 [7]3 years ago
4 0

Answer:

D: It is unlikely that a specific cause can be determined, but the treatment would likely be the same in either case.

Explanation:

Taking exam right now and got it correct

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What is veces??????????????????????????????????
tangare [24]
Veces is Spanish word for times.Hope I helped.
3 0
3 years ago
Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the
rusak2 [61]

Answer:

Option d: C₈H₉NO₂ = acetaminophen, analgesic

Explanation:

% composition of compound is:

63.57 g of C

6 g of H

9.267 g of N

21.17 g of O

First of all we divide each by the molar mass of the element

63.57 g / 12 gmol = 5.29 mol of C

6 g of H / 1 g/mol = 6 mol H

9.267 g of N / 14 g/mol =  0.662 mol of N

21.17 g of O / 16 g/mol = 1.32 mol of O

We divide each by the lowest value, in this case 0.662

5.29 / 0.662 = 8

6 / 0.662 = 9

0.662 / 0.662 = 1

1.32 / 0.662 = 2

Molecular formula of the compound is C₈H₉NO₂

7 0
3 years ago
HELPP! pls
pshichka [43]

Answer:

"2.48 mole" of H₂ are formed. A further explanation is provided below.

Explanation:

The given values are:

Mole of Al,

= 3.22 mole

Mole of HBr,

= 4.96 mole

Now,

(a)

The number of mole of H₂ are:

⇒  \frac{Mole \ of \ H_2}{3} =\frac{Mole \ of HBr}{6}

or,

⇒  Mole \ of \ H_2=\frac{1}{2}\times Mole \ of \ HBr

⇒                      =\frac{1}{2}\times 4.96

⇒                      =2.48 \ mole

(b)

The limiting reactant is:

= HBr

(c)

The excess reactant is:

= Al

6 0
3 years ago
An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0
3 years ago
Which of the following is an example of a chemical change? A. melting solid gold B. burning hydrogen gas C. breaking a sheet of
cestrela7 [59]
The answer is B. Burning hydrogen gas

5 0
3 years ago
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