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OLEGan [10]
3 years ago
8

Please help me with this! I’ll mark you brainiest if it’s correct. I do know it’s either A or C

Chemistry
2 answers:
8090 [49]3 years ago
8 0
It is A
I only know that Lithium does not have Roman numerals after it only certain elements do so yea!
Volgvan3 years ago
6 0

Answer:

the correct answer is A

Explanation:

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The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x
Andrews [41]
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0
3 years ago
A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is
makvit [3.9K]

Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

\frac{V}{T} =k

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

\frac{V1}{T1} =\frac{V2}{T2}

<h3>Temperature of the gas in this case</h3>

In this case, you know:

  • P1= 1800 psi
  • V1= 10 L
  • T1= 20 C= 293 K (being 0 C= 273 K)
  • P2= 1800 psi
  • V2= 6 L
  • T2= ?

You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

\frac{10 L}{293 K} =\frac{6 L}{T2}

Solving:

\frac{10 L}{293 K} T2=6 L

T2=\frac{6 L}{\frac{10 L}{293 K} }

<u><em>T2=163.8 K= -109.2 C</em></u>

The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

brainly.com/question/4147359?referrer=searchResults

7 0
3 years ago
g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empir
sammy [17]

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

7 0
3 years ago
The formula ca(no3)2 tells us that one formula unit of this compound is composed of ________ calcium atoms, ________ nitrogen at
ad-work [718]

Answer:- The formula Ca(NO_3)_2 tells us that one formula unit of this compound is composed of one calcium atom, two nitrogen atoms, and six oxygen atoms.

Explanations:- Subscripts tell us about the number of atoms of the element for which they are used. For example, here the subscript of Ca is one, it means there is one calcium atom in the given one formula unit.

When we have subscripts inside and outside the parenthesis then they are multiplied and the outside subscript is considered for all the atoms present inside the parenthesis.

Here, for the given chemical formula, the subscript of N is 1 and the subscript present outside is 2. So, 1 x 2 = 2 and for oxygen, 3 x 2 = 6

So, we have one calcium atom, two nitrogen atoms and six oxygen atoms for one formula unit of given compound.

3 0
3 years ago
Read 2 more answers
What is the percent by mass of oxygen in ca(oh)2? [formula mass = 74.1]
andre [41]
16(2)/74.1=.431

.431x100= 43.1%
8 0
4 years ago
Read 2 more answers
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