Ask question

Ask question

All categories

2 years ago

14

6. Which of the following labels for the picture below is correct? a) Solute + solvent = solution b) Solvent + solute = saturati

on c) Solvent + solute = soluble d) Solvent + solute = solution

1 answer:

SVETLANKA909090 [29]2 years ago

4 0

Answer:

answer is a since solute dissolve a solvent to give a solution

You might be interested in

The molality is temp independent whereas molarity is temperature dependent. Justify.

Elza [17]

Molality is obtained by dividing the number of moles of solute by the mass in kilogram of the solvent. None of the dimensions is dependent in temperature. On the other hand, molarity is obtained by dividing the number of moles of solute by the volume in liters of the solution. Volume is temperature dependent.

6 0

3 years ago

Which inorganic substance taken up by plants is also taken in and used directly by animals, including humans?

masha68 [24]

Answer: water

water can not be organic...but is is took by humans and plants because theur life dpends on it

Explanation:

3 0

3 years ago

Read 2 more answers

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

3 years ago

A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity

grandymaker [24]

Answer:

Mass of Ca in sample, Mass of Br in sample, Number of moles of Ca in sample, Number of moles of Br in sample, Mass or moles of element other than Ca or Br in sample

Explanation:

The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.

Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

8 0

2 years ago

A Carnot cycle operates between the temperatures limits of 400 K and 1600 K, and produces 3600 kW of net power. The rate of entr

TiliK225 [7]

The rate of entropy change:

The rate of entropy change of the working fluid during the heat addition process is 3 kW/K

What is the Carnot cycle?

The Carnot Cycle is a thermodynamic cycle made up of reversible isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression processes in succession.
The ratio of the heat absorbed to the temperature at which the heat was absorbed determines the change in entropy.

The entropy of a system:

The rate of heat addition is expressed as,

Q = $\frac{WT_{H}}{T_{H}- T_{L}}$

The entropy of a system is a measure of how disorderly a system is getting. The rate of entropy generation during heat addition is,

$S_{gen} = \frac{Q}{T_{H}} = \frac{W}{T_{H} - T_{L}}$

Calculation:

<u>Given:</u>

$T_{L}$ = 400K

$T_{H}$ = 1600K

W = 3600 kW

Put all the values in the above equation, and we get,

$S_{gen} = \frac{W}{T_{H} - T_{L}}$ = $\frac{3600}{1600-400}$ = 3 kW/K

The rate of entropy change is 3 kW/K

Learn more about the Carnot cycle here,

brainly.com/question/13002075

#SPJ4

3 0

1 year ago