A solution is dilute when?A. it has a lot of solute
B. has little solvent
C. it has a lot of solvent
D. has maximum solute

Answers

Answer 1
Answer: The answer to your equation is C. Hope this Helps!
Answer 2
Answer: C. It has more solvent
To dilute a solution means to add more solvent without the addition of more solute

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The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.
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Draw the Lewis structure for XeCl2 and answer the following questions.How many valence electrons are present in this compound?
How many bonding electrons are present in this compound?
How many lone pair (non-bonding) electrons are present in this compound?

Answers

Answer:

Valence electrons in XeCl2 = 8 + 7 + 7 = 22.

Bonding electrons = 4.

Nonbonding electrons = 18.

Explanation:

Hello.

In this case, you can see the Lewis structure on the attached picture, in which you can see that there are since xenon has 8 valance electrons and each chlorine has 7 valence electrons, the total amount of valence electrons is:

Valence electrons in XeCl2 = 8 + 7 + 7 = 22.

Moreover, since each chlorine atom is bonding with one of the eight electrons of xenon (Lewis structure), we can see there are 4 bonding electrons.

Finally, since there are six nonbonding electrons per chlorine atom and six nonbonding electrons in xenon, the overall nonbonding electrons are:

Nonbonding electrons in XeCl2 = 6 + 6 + 6 = 18.

Regards.

A mass of 34.05 g of H2O(s) at 273 K is dropped into 185 g of H2O(l) at 310. K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equi- librium has been reached. Assume that CP, m for H2O is con- stant at its values for 298 K throughout the temperature range of interest.

Answers

Answer:

The temperature of the system once equilibrium is reached, is 292 Kelvin

Explanation:

Step 1: Data given

Mass of H2O = 34.05 grams  

⇒ temperature = 273 K

Mass of H2O at 310 K = 185 grams

Pressure = 1 bar = 0.9869 atm

Step 2: Calculate the final temperature

n(ice)*ΔH(ice fusion) + n(ice)*CP(H2O)(Tfinal- Ti,ice) + n(H20)*CP(H2O)*(Tfinal-Ti,H2O) = 0

Tfinal = [n(ice)*CP(ice)*Ti(ice) + n(H2O)*CP(H2O)*Ti(H20) - n(ice)*ΔH(ice fusion)] / [n(ice)*CP(ice) +n(H2O)*CP(H2O)]

⇒ with n(ice) = moles of ice = 34.05 grams / 18.02 g/mol = 1.890 moles

⇒ with CP(ice) = 75.3 J/K*mol

⇒ with Ti(ice) = the initial temperature of ice = 273 K

⇒ with n(H2O) = the moles of water = 185.0 grams / 18.02 g/mol = 10.27 moles

⇒ with CP(H2O) = CP(ice) = 75.3 J/K*mol

⇒ with Ti(H2O) = the initial temperature of the water = 310 K

⇒ with ΔH(ice, fusion) = 6010 J/mol

Tfinal = [1.890 moles * 75.3 J/K*mol * 273 + 10.27 mol * 75.3 J/K*mol * 310 K - 1.890 moles * 6010 J/mol] / [1.890 moles *75.3J/k*mol + 10.27 mol * 75.3 J/K*mol]

38852.541 + 239732.61  - 11358.9 = 267226.251

Tfinal= 291.8 ≈ 292 Kelvin

The temperature of the system once equilibrium is reached, is 292 Kelvin

Final answer:

We must use the principle of conservation of energy to equate the heat gained by the ice to the heat lost by the water, in a given equation, to calculate the final equilibrium temperature.

Explanation:

The topic here is thermodynamics, specifically calculating the final equilibrium temperature when two substances are mixed. Given the information, we can apply the principle of conservation of energy, which in this context is the heat gained by one substance is equal to the heat lost by the other. In this case, the heat gained by the ice (H2O(s)) as it melts and increases in temperature is equal to the heat lost by the water (H2O(l)). Therefore, we have the equation 34.05 g * 1 kcal/kg * K *(T - 273 K) + 34.05 g * 80 Cal/g = 185 g * 1 kcal/kg*K *(310 K - T) where T is the final temperature to be solved.

Learn more about heat transfer here:

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Which of the following examples illustrates a number that is correctly rounded to three significant figures? a. 109 526 g to 109 500 g
b. 0.03954 g to 0.040 g
c. 20.0332 g to 20,0 g
d. 04.05438 g to 4,054 g
e. 103.692 g to 103.7g

Answers

Answer:

c. 20.0332 g to 20,0 g

Explanation:

A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.

Which of the following examples illustrates a number that is correctly rounded to three significant figures?

a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.

b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures:  0.040.

c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures:  20.0.

d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.

e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.

A substance that cannot be chemically broken down into simpler substances is a an electron. b a heterogeneous mixture. c an element. d a homogeneous mixture. e a compound.

Answers

Answer:

c. an element.

Explanation:

An element -

It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .

In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .

Hence , from the given information of the question ,

The correct term is an element  .

Answer:

C. an element.

Explanation:

During an experiment, a student adds 2.90 g CaO to 400.0 mL of 1.500 M HCl . The student observes a temperature increase of 6.00 °C . Assuming that the solution's final volume is 400.0 mL , the density is 1.00 g/mL , and the heat capacity is 4.184 J/g⋅°C , calculate the heat of the reaction, ΔHrxn .

Answers

Answer:

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl  → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture

Answers

There is 65% of NaHCO3 in the sample.

The equation of the reaction is;

HA + NaHCO3 -----> NaA + CO2 + H2O

Amount of CO2 formed = mass/molar mass

mass of CO2 = 0.561 g/44 g/mol = 0.013 moles

From the balanced reaction equation;

1 mole of NaHCO3 yields 1 mole of CO2

0.013 moles of Na2CO3 yields 0.013 moles of CO2

Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3

Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1

= 65%

Learn more: brainly.com/question/25150590

Answer:

63.75%.

Explanation:

The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;

HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).

Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.

Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.

Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.

Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.

Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.