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Svet_ta [14]
3 years ago
6

what is meant by a calibration curve for a colored substance being studied with a spectrophotometer? How is a calibration curve

used to determine the concentration of a solution of unknown concentration of the colored substance?
Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
3 0

Answer:

The concentration of the analyte is determined by fitting the absorbance or transmittance obtained by spectrophotometric analysis of the unknown solution into the calibration curve.

Explanation:

In a calibration curve, the instrumental response (absorbance or transmittance), is plotted against the concentration of the analyte (the substance to be measured). The analyst is expected to prepare a series of standard solutions of the analyte within a range of solution concentrations close to the expected concentration of analyte in the unknown solution. The method of least squares may be used to determine the best fit of the line, thus, the concentration of the analyte. This method is only used for the determination of the concentration of coloured substances (spectrophotometry).

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Please I need help with questions 5-9 and it’s very hard and I’m struggling with it and if you need to see the picture big then
Serggg [28]
"the answer will be 0.00 "
3 0
3 years ago
If 8.6 g of ch4 and 5.9 g of o2 react, what is the mass, in grams, of h2o that is produced?
Alenkasestr [34]
This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water. 

REMEMBER, you have to write and balance the chemical equation before you can do any of that work. 
That happens to be CH4 + 2O2 => CO2 + 2H2O
7 0
3 years ago
If 1.76 g of an ideal gas occupy 1.0 L at standard temperature and pressure (STP), what is the molar mass of the gas?
ycow [4]

Answer:

Explanation:

Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :) 

8 0
2 years ago
3. The temperature of a fixed mass of gas in a rigid container is decreased from 127.00 degrees
Sonbull [250]

Answer:

750mmHg

Explanation:

The following data were obtained from the question:

T1 = 127°C = 127 +273 = 400K

T2 = 27°C = 27 +273 = 300K

P1 = 1000mmHg

P2 =?

P1/T1 = P2/T2

1000/400 = P2 /300

Cross multiply

400 x P2 = 1000 x 300

Divide both side by 400

P2 = (1000 x 300)/400

P2 = 750mmHg

Therefore, the new pressure after cooling is 750mmHg

5 0
3 years ago
Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

7 0
1 year ago
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