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3 years ago

How could you ensure that all the spectroscopes created the same calibration line and equation?

2 answers:

8 0

In order to ensure that all the spectrosocpes created the same calibration line and equation, one must follow National Institute of Standards and Technology (NIST) standards for comparison.

NIST is authorized to issue certificate for calibration which certifies that the calibration line and equation are upto the standards decided.

alekssr [168]3 years ago

3 0

How could you ensure that all the spectroscopes created the same calibration line and equation?

Answer: You would simply have to take all the spectroscopes reading and ensure that all the helium lines are recorded in regard with the scale along with their colors and wavelengths. All this data should be graphed on a calibration plot and analyzed.

I hope it helps, Regards.

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zzz [600]

Answer:

i think kinetic energy sorry if I'm wrong.

6 0

3 years ago

1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a

Archy [21]

Answer:

Explanation:

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃ + Al → 3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

Al : Ag

1 : 3

0.223 : 3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂ → 2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

HgO : Hg

2 : 2

3.12 : 3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂ + 5O₂ → 2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

N₂O₅ : O₂

2 : 5

12.99 : 5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

C₆H₆ : CO₂

2 : 12

0.103 : 12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0

3 years ago

A gas has a volume of 20.0 L at a pressure of 950 mmHg with a temperature of 125.0 o If the conditions are changed to STP, what

ehidna [41]

Use formula: Initial Pressure x Initial Volume/Initial temperature = Final pressure x Final Volume/Final Temperature => 17.15L

6 0

3 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

6 0

2 years ago

How many grams of sulfur are needed to react completely with 246 g of mercury to form hgs?

iren [92.7K]

The reaction between mercury (Hg) and sulfur (S) to form HgS is:

Hg + S ------------- HgS

Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS

The given mass of Hg = 246 g

Atomic mass of Hg = 200.59 g/mol

# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles

Based on the reaction stoichiometry,

# moles of S that would react = 1.226 moles

Atomic mass of S = 32 g/mol

Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g

39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS

5 0

2 years ago

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