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Object A has a molar heat of 31.2 J/mole∙°C and object B molar heat is 11.2 J/mole∙°C. Which object will heat up faster if they have the same mass and equal amount of heat is applied? Explain why.

Answers

Answer 1

Answer:

Answer:

Substance B

Explanation:

Molar heat of A = 31.2J/mole.°C

Molar heat of B = 11.2 J/mole∙°C.

The molar heat of a substance is the amount of heat that must be added to a mole of a substance to raise the temperature by 1°C.

Substance B will heat up faster compared to A.
It has a smaller molar heat compared to A.
This suggests that it will require lesser heat to raise its temperature by 1°C.

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22. What is the molar mass of oxygen (O2)?A- 15.9994 g/molB- 1.204 x 1024 g/molC- 6.02 x 1023 g/molD- 31.9988 g/mol
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Calculate the molality of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4) (35.4 % means 35.4 g of H3PO4in 100 g of solution)

Answers

Answer:

3.6124 m/kg

Explanation:

Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us 35.4 g of H3PO4 and 100 g of solution to work with.

To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.

Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:

35.4 g x 1 mol / 97.9952 g = 0.3612422 mol

Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.

To find the molality, divide the moles of solute and kilograms of solution.

0.3612422 mol / 0.1 kg = 3.6124 m/kg

A lab group was calculating the speed of a radio car. They measured the distance traveled to be 6 meters and the time to be 3.5 seconds. Then they divided the distance by the time to find the speed. The actual speed was 2.2 m/s. What was their percent error?

Answers

Percent error is a measure of the difference between an observed value and a true value.

Actual Speed (True Value) = 2.2 m/s

Experimental Speed (Calculated Value) = Distance / Time = 6 m / 3.5 s = 1.714 m/s

The formula for calculating percent error is:

Percent Error = ((|Actual Value - Experimental Value|) / |Actual Value|) * 100%

Calculate the absolute difference between the actual speed and the experimental speed:

|2.2 - 1.714| = 0.486

Calculate the absolute value of the actual speed:

|2.2| = 2.2

Percent Error = (0.486 / 2.2) * 100%

= 0.221 * 100%

= 22.1%

The calculated percent error is approximately 22.1%. This means that the lab group's calculated speed of 1.714 m/s is about 22.1% lower than the true speed of 2.2 m/s.

Percent error is a way to quantify the accuracy of experimental measurements. A positive percent error indicates that the experimental value is higher than the true value, while a negative percent error indicates that the experimental value is lower. In this case, since the calculated speed is lower than the true speed, we have a positive percent error.

To know more about actual speed here

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Answer:456

Explanation:

Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together. Which statement is correct? a.Both NaCl and CuS will precipitate from solution b.No precipitate will form c.Only CuS will precipitate from solution d.Only NaCl will precipitate from solution

Answers

C.Only CuS will precipitate from solution

Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together.

According to question theEquation :

Na₂S(aq) + CuCl₂(aq) → Products

The complete balanced reaction is:

Na₂S(aq) + CuCl₂(aq) → CuS(s) ↓ + 2NaCl(aq)

When the sulfide bond to the cation Cu²⁺ it makes a precipitate (s)

Thus,the correct answer is C.

Learn more :

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Answer:

The correct state for the answer is c.

Only CuS will precipitate from solution

Explanation:

We analyse the compounds for the reaction and we write the equation:

Na₂S(aq) + CuCl₂(aq) → Products

For this case, the products are NaCl and CuS. The complete balanced reaction is:

Na₂S(aq) + CuCl₂(aq) → CuS(s) ↓ + 2NaCl(aq)

When the sulfide bond to the cation Cu²⁺ it makes a precipitate (s)

Salts from chlorides are soluble except for the Ag⁺, Pb⁺ or Cu⁺

Salts from S⁻² which are soluble, are found in the group 2 of the Periodic Table (Ca²⁺, Ba²⁺, Mg²⁺)

The correct state for the answer is c.

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255?

Answers

Answer is 231.8 K

Explanation
We can use Charle's law to solve this problem.

Charle's law says"at a constantpressure, the volume of a fixed amount of gas is directly proportional to itsabsolute temperature".

V α T

Where V is the volume and T is the temperaturein Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 1.00 L
T₁ =?
V₂ = 1.10 L
T₂ = 255 K (Since the unit of the given temperature is missing in the question, thought that the given temperature is in Kelvin)

By applying theformula,
1.00 L / T₁ = 1.10 L / 255 K
T₁ =(1.00 L / 1.10 L) x 255 K
T₁ = 231.8 K

Hence, the initial temperature is 231.8 K.

Assumption : the pressure of the gas is a constant.

Hello!

What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255 ?

We have the following data:

V1 (initial volume) = 1.00 L

V2 (final volume) = 1.10 L

T1 (initial temperature) = ? (in Kelvin)

T2 (final temperature) = 255 K

According to the Law of Charles and Gay-Lussac in the study of gases, in an isobaric transformation, ie when a mass under pressure maintains its constant pressure, on the other hand, as the volume increases, the temperature increases and, if the volume decreases, the temperature decreases (directly proportional to temperature and volume) . We apply the data to the formula of isobaric transformation (Charles and Gay-Lussac), we will see:

$(V_1)/(T_1) = (V_2)/(T_2)$

$(1.00)/(T_1) = (1.10)/(255)$

multiply the means by the extremes

$1.10*T_1 = 1.00*255$

$1.10\:T_1 = 255$

$T_1 = (255)/(1.10)$

$\boxed{\boxed{T_1 \approx 231.8\:K}}\:\:\:\:\:\:\bf\red{\checkmark}$

Answer:

The initial temperature is approximately 231.8 Kelvin

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Energy levels in an electron configuration correspond to what on the periodic table?

Answers

I believe energy levels correspond to periods on the periodic table.

How many grams of sucrose must be added to 375 mL of watertoprepare a 2.75m/m percent solution of sucrose?

Answers

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 375 mL

$\text{Mass of water}=1.00g/mL* 375mL=375g$

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain $(375)/(97.25)* 2.75=10.6$ grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

Density of water ≈ 1 g/mL

Mass of water = Volume of water × Density of water

Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

Mass of sucrose = (Desired m/m percent / 100) × Total mass

Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))