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Ivanshal [37]
3 years ago
12

How could you ensure that all the spectroscopes used the same calibration line and equation? What would you need to be sure was

always the same? What could be changed without affecting the calibration?
Chemistry
1 answer:
NISA [10]3 years ago
8 0
People had asked this many times and that is why they came up with methods and standards that will answer these type of questions. You can look it up in the NIST or the  National Institute for Standards and Technology.
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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
PLZ HELP GOD PLZ HELP ME PLZ AWNSER 11. AND 14.
JulsSmile [24]
  1. 11, ions
  2. 14, electrolyte

hope this helps

6 0
3 years ago
Write a balanced chemical equation, including states of matter, for the combustion of gaseous benzene, c6h6.
snow_tiger [21]
Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air. 

<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
4 0
3 years ago
How many atoms are in 5.70 x 10^32 mol of Rn?
Alex73 [517]
To find amount of atoms from mol, multiply the mole amount by Avogadro’s number

5.70x10^32 x 6.02x10^23

= 3.43x10^56 atoms
5 0
3 years ago
10) Which gas sample at STP has the same number of molecules as a 2.0 Liter
goblinko [34]

Answer:

The sample at STP that has the same total number of molecules as 2.0 liters of CO2(g) at STP is 2) 2.0 L of CI2 (g) New questions in Chemistry Exercise 1-8 Calculate the number of moles of 4.82 x 10 to the power of 24 iron atoms

Explanation:

7 0
3 years ago
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