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Mumz [18]
3 years ago
9

Visible light waves are part of the blank Spectrum

Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0

Visible light spectrum or electromagnetic spectrum

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A percent composition analysis yields 46.7% nitrogen and 53.3% oxygen. What is the empirical formula for the compound
Korolek [52]
Empirical formula is the simplest ratio of whole numbers of components in a compound
calculating for 100 g of compound 
                                                     N                                   O
mass                                          46.7 g                              53.3 g
number of moles                  46.7 g / 14 g/mol                 53.3 g / 16 g/mol 
                                                  = 3.33                               = 3.33
divide by the least number of moles 
                                                3.33/ 3.33 = 1.00              3.33/3.33 = 0
number of atoms 
N - 1
O - 1
therefore empirical formula is NO
6 0
3 years ago
A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. T
zubka84 [21]

Answer: The moles of gas present in the cylinder is 0.34 moles.

Explanation:

Given: P_{1} = 2.7 atm,   V_{1} = 3.1 L,     T_{1} = 300 K

P_{2} = ?,      V_{2} = 9.4 L,       T_{2} = 610 K

Formula used to calculate the final temperature is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm

Now, moles present upon heating the cylinder are as follows.

P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol

Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.

6 0
3 years ago
A block of dry ice (-40°C) is placed in contact with an ice cube (-10°C).
const2013 [10]

Explanation:

Answer is that : no heat will flow

3 0
3 years ago
The alkali metals all react with water. which is the most reactive metal?
kifflom [539]
Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be FRANCIUM, which is at the base of Group One.


Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.

8 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

6 0
3 years ago
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