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AVprozaik [17]
3 years ago
13

How do I do this whole problem?

Chemistry
1 answer:
KatRina [158]3 years ago
4 0

The solubility of a substance in water is dependent on the temperature. Thus for

1 & 2: Temperature is the independent variable (the one that changes in the first place) and Solubility is a dependent variable (a variable that changes in response to changes in the independent variable.)

The graph: by convention you shall label the horizontal axis with the independent variable and the vertical axis with the dependent variable. For clarity's sake you shall use the finest scale possible that accommodates for all data provided for both axis. Plot the data points on the graph as if they are points on a cartesian plane.

My teacher made no detailed requirements on the phrasing on titles of solubility curve plots; however, like most other graphs in chemistry, the title shall specify the name of variables presented in this visualization. For instance, "the solubility of KClO_3 under different temperatures" might do. You shall refer to your textbooks for such convention.

It is necessary to interpolate to find the solubility at a temperature not given in the table. Start by connecting all given data points with a smooth line; find the vertical line corresponding to temperature = 75 degree Celsius and determine the solubility at the intersection of the vertical line and the trend line. That point shall approximates the solubility of the salt at that temperature.

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Answer:

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4 0
2 years ago
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C + 2H2 -> CH4
ankoles [38]

Answer:

<u>C) 4</u>

Explanation:

<u>The reaction</u> :

  • C (s) + 2H₂ (g) ⇒ CH₄ (g)

       12g      4g             16g

Hence, based on this we can say that : <u>2 moles of hydrogen gas are needed to produce 16g of methane.</u>

<u />

<u>For 32g of methane</u>

  • Number of moles of H₂ = 32/16 × 2
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4 0
1 year ago
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g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
2 years ago
A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom
Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

(quit)

polyalchemVirtuoso

Answer:

Explanation:

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