Directions: Find the scale for and volume shown in each graduated cylinder. You should estimate the final place.

How many kj of heat are needed to completely melt 32.3 g of h2o, given that the water is at its melting point? the heat of fusio

timurjin [86]

Answer: fourth option, 10.8 kJ

Explanation:

The heat of fusion, also named latent heat of fusion, is the amount of heat energy required to change the state of a substance from solid to liquid (at constant pressure).

The data of the heat of fusions of the substances are reported in tables and they can be shown either per mole or per gram of substance.

In this case we have that the heat of fusion for water is reported per mole: 6.02 kJ/mole.

The formula to calculate how many kJ of heat (total heat) are needed to completely melt 32.3 g of water, given that the water is at its melting point is:

Heat = number of moles × heat of fusion

The calculations are:

number of moles = mass / molar mass

number of moles = 32.3 g / 18.015 g/mol = 1.79 mol

Heat = 1.79 mol × 6.02 kJ / mol = 10.8 kJ ← answer

5 0

3 years ago

Read 2 more answers

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

Find the no. Of electron involved in the electro deposition of 63.5g of cu from a solution of cuso4

sp2606 [1]

This is a redox reaction, meaning reduction-oxidation reaction. This represents the reaction in one side of the electrode in an electrolysis set-up. First, we find the oxidation number of Cu in CuSO4:

(ox. # of Cu)+ ox.# of S + 4(ox.# of oxygen) = 0
(ox. # of Cu) + (6) + 4(-2) = 0
ox. # of Cu = 2+

CuSO4 ---> Cu + SO42-
Cu2+ + SO42- ----> Cu + SO42-
Cu2+ -----> Cu + 2e- (net ionic reaction)

The stoichiometric equation would be 2 electrons per mole Copper. Copper has a molar mass of 63.5 g/mol. Then, it would only need 2 electrons.

8 0

3 years ago

You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph

zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

C1V1 = C2V2

where

C1 is the concentration of stock solution
V1 is the volume of stock solution required to prepare a diluted solution
C2 is the concentration of the diluted solution prepared
V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0

2 years ago

Contrast the way the conqueror comes with the way the Pilgrims came (stanzas three and four).

Masja [62]

Answer:

tbh id k i've never read it so if u could send the link I could help you.

Explanation:

7 0

3 years ago