Answer:
1.99 x 10⁻¹⁸J
Explanation:
Given parameters:
Frequency of the wave = 3 x 10¹⁵Hz
Unknown:
Energy of the photon = ?
Solution:
To solve this problem, we use the expression below;
E = hf
Where E is the energy, h is the Planck's constant and f is the frequency
Now insert the parameters and solve for E;
E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J
The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
Learn more about empirical formula at: brainly.com/question/14044066
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Answer:
Ratio is 1:1
Explanation:
I do not see any coefficients infront of the reactants and the products, therefore, we can automatically assume that every reactant and product is 1 mole. Don't get confused by the 4 off the O. It just means that 1 mole of sulfate has 1 zinc and 4 oxygens.
Answer:
Sedimentary only
Explanation:
when compacting and cementing happens then a sedimentary rock can and is made!
hope this helps!
Answer:
A
Explanation:
To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.
We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.
Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.
