Ask question

Ask question

All categories

3 years ago

13

If 225 cal of heat is added to 15.0 g of 2-propanol at room temperature, 25ºC, the alcohol will reach a final temperature of ___

_____ ºC. The specific heat of 2-propanol is 0.612 cal/(g•ºC).

1 answer:

natima [27]3 years ago

6 0

Explanation:

I have no idea about this

You might be interested in

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

You will be reading from an essay and a play dealing with the lives of women in the late eighteenth and early

s344n2d4d5 [400]

Answer:

all work

Explanation:

6 0

3 years ago

Read 2 more answers

When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, h

Pavel [41]

NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>

6 0

3 years ago

How many formula units are present in 40.0 grams of Cu2S?

11Alexandr11 [23.1K]

Never mind, I did the problem wrong, I deeply apologize.

4 0

2 years ago

How many moles of hydrogen are required to react with 4.6 x 10 22 molecules of nitrogen?

V125BC [204]

Answer: 6 moles

Take a look at the balanced chemical equation for this synthesis reaction

N 2(g] + 3 H 2(g] → 2 NH 3(g]

Notice that you have a 1:3 mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.

So, if you have 2 moles of nitrogen taking part in the reaction, you will need

2 moles N 2 ⋅ 3 moles H 2 /1 mole N 2 = 6 moles H 2

8 0

3 years ago