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Romashka [77]
3 years ago
13

What is the percent composition of Br in CuBr3?

Chemistry
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

about 79% (79.04369332 to be exact)

Explanation:

Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100

Br= 3 x 79.9/303.25 x100=79.04369332

You might be interested in
explain the 3 ways the elements were formed and which elements were formed in each method. (Big Bang, Stellar, And Supernova Nuc
svp [43]

Answer:

See explanation

Explanation:

Big Bang: During the big bang, elements were formed by the squeezing together of light nuclei. Hydrogen atoms were fused together to make helium,  lithium  and other light weight elements.

Stellar Nucleosynthesis: Stellar nucleosynthesis is the process by which elements are created within stars by combining the protons and neutrons together from the nuclei of lighter elements(Thoughtco). In the stars, hydrogen may be fused to give helium, lithium and other elements lighter than iron.

Supernova Nucleosynthesis: A supernova is a process in which a star releases a large amount of energy together with neutrons, hence elements that are heavier than iron, like uranium and gold, can be produced in the process.

4 0
3 years ago
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is
WARRIOR [948]

The given question is incomplete, the complete question is:

A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 18.0cm wide and 21.6cm high. The maximum safe pressure inside the vessel has been measured to be8.50MPa For a certain reaction the vessel may contain up to0.362kg of carbon dioxide gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius.Be sure your answer has the correct number of significant digits.

Answer:

The correct answer is 411 degree C.

Explanation:

The width of the vessel is 18.0 cm or 9.0 cm in radius and height of the vessel is 21.6 cm. In the given question, to find the volume of the cylinder, the formula to be used is pie r^2h

= 22/7 * (9)^2 * 21.6 cm^3

= 5498.7 cc

= 5.5 L

The value of 1MPa is 9.869 atm, therefore, the value of 8.5 MPa will be 83.89 atm.  

The maximum safe operating temperature can be calculated by using the ideal gas equation, that is, P = nRT/V or T = PV/nR

= 83.89 atm * 5.5 L / (0.362*10^3 g/44g/mol) * 0.082 L atmK^-1mol^-1

= 683.9 K

= 683.9 - 273 = 410.9 degree C = 411 degree C.  

Hence, the maximum safe operating temperature is 411 degree C.  

6 0
4 years ago
Which of the following is most likely to have a crystalline structure
uranmaximum [27]
Ionic bonds are most likely solid crystalline at room temperature
7 0
3 years ago
Given the speed of light as 3.0 × 108 m/s, calculate the wavelength of the electromagnetic radiation whose frequency is 7.5 × 10
arlik [135]
Wavelength= velocity/frequency
wavelength= (3.0 x 10^8m/s) / 7.5 x 10^12 Hz)
you can do the math
I am assuming u that 108 is 10^8 and the 1012 is 10^12
3 0
4 years ago
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes
Lisa [10]

Answer:

pH = 8.72

Explanation:

This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

As this compound acts like a base, we propose this equilibrium:

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰

Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M

So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

We can avoid the quadractic equation because Kb is so small

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH = 8.72

4 0
3 years ago
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