You to cute ✌✌<br><br> i will give u brailtist if u send me something funny.

a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

Diano4ka-milaya [45]

<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

= (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

= (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

X = 32.978°C

= 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0

4 years ago

A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,

lara31 [8.8K]

Answer:

The answer to your question is: T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L V2 = 2.78 L

P1 = 2.40 atm P2 = 1.97 atm

T1 = 325°K T2 = ?

Formula

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

Process

T2 = (P2V2T1) / (P1V1)

T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

T2 = 1779.895 / 7.56

T2 = 235.44 °K

4 0

3 years ago

1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour

lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

5 0

3 years ago

Calculate the standard free-energy change at 25 ∘C for the following reaction:

lianna [129]

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$ ( $\Delta G^{0}$ ): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

8 0

3 years ago

Read 2 more answers

Conversion factors:

pantera1 [17]

The number of micrograms of contaminant each person will receive is 5,742.64 ug.

<h3>Number of micrograms of contaminant per person</h3>

The number of micrograms of contaminant each person will receive is calculated as follows;

Amount of contaminant (mg) = (1.01 x 10⁻⁴ mg/mL) x 2.28 x 10⁴ L

Amount of contaminant (mg) = 2.3028 mg.L/mL

Amount of contaminant (mg) = (2.3028 mg.L/mL) x (1000 mL/L) = 2,302.8 mg

Amount of contaminant (ug) = 2,302.8 mg x 1000 ug/mg = 2,302,800 ug

<h3>Amount of contaminant per person (ug/person) </h3>

= (2,302,800 ug) / (401 persons)

= 5,742.64 ug per person

Learn more about micrograms of contaminant here: brainly.com/question/14522199

#SPJ1

8 0

2 years ago

You to cute ✌✌ i will give u brailtist if u send me something funny.