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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

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Match the vocabulary terms to their definitions.
mariarad [96]

Answer:

Units Accuracy = the degree to which a measurement can be

Column Matter = anything that has mass and occupies

Space Precision = an indication of how close a measurement is to the correct

Replicated Meniscus = the curved top surface of a liquid

Result Density = the mass of an object per unit volume

Volume = space occupied, measured in cubic

Explanation:

In the fields of science , the Unit Accuracy of a measurement system is the degree of closeness of measurements of a quantity to that quantity's true value.

Column Matter can be defined as anything that had mass and occupies space.

Space Precision refers to the closeness of the measurements to each other. It's the quality, condition, or fact of being exact and accurate.

Replicated Meniscus is the curve in the upper surface of a liquid close to the surface of the container or another object, caused by surface tension. It can be either concave or convex, depending on the liquid and the surface.

Result Density can be defined as the mass of an object per unit volume.

Volume: The space occupied by any object is called the volume. The volume of an object is the amount of space occupied by the object. Volume is measured in "cubic units".

8 0
3 years ago
you find this receipt and remember it can help you with a lock. You have 9 members in your unit. This will take care of 1 member
Stells [14]

Answer:(9x) x 6 + (8x) x= the cost to take care of one member for one day

Explanation:

the receipt takes care of one of the members for one day which is the "+ (8x)" there is no number to represent the cost, so this is all that's possible

7 0
3 years ago
A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o
koban [17]

Answer:

CH₅N

Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

(45.0 g)(mol/18.02g) = 2.497...mol H₂O

Moles of H is found using the molar ratio of 2H:1H₂O:

(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

3 0
3 years ago
Butane reacts with oxygen to produce carbon dioxide and water
Wittaler [7]
What exactly are you looking for?
This is the balanced equation.
<span>2 C4H10g + 13 O2g ---> 8 CO2g + 10 H2Og</span>
4 0
3 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
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