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3 years ago

Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ

Chemistry

1 answer:

masya89 [10]3 years ago

5 0

Answer:

endet nach selam nw

4gh7

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In a constant‑pressure calorimeter, 70.0 mL of 0.770 M H2SO4 is added to 70.0 mL of 0.420 M NaOH. The reaction caused the temper

schepotkina [342]

Answer : The enthalpy of neutralization is, -113.9 KJ/mole

Explanation :

First we have to calculate the moles of $H_2SO_4$ and $NaOH$ .

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.770mole/L\times 0.070L=0.0539mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.420mole/L\times 0.070L=0.0294mole$

The balanced chemical reaction will be,

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 2 mole of $NaOH$ neutralizes by 1 mole of $H_2SO_4$

As, 0.0294 mole of $NaOH$ neutralizes by $\frac{0.0294}{2}=0.0147$ mole of $H_2SO_4$

Thus, the number of neutralized moles = 0.0147 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 = $70.0ml+70.0ml=140.0ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 140.0ml=140.0g$

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 = $4.184J/g^oC$

m = mass of water = 140.0 g

$T_{final}$ = final temperature of water = $26.00^oC$

$T_{initial}$ = initial temperature of metal = $23.14^oC$

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

$q=140.0g\times 4.184J/g^oC\times (26.00-23.14)^oC$

$q=1675.27J=1.675kJ$

Thus, the heat released during the neutralization = -1.675 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -1.675 KJ

n = number of moles used in neutralization = 0.0147 mole

$\Delta H=\frac{-1.675KJ}{0.0147mole}=-113.9KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, -113.9 KJ/mole

3 0

3 years ago

Please I need help with questions 5-9 and it’s very hard and I’m struggling with it and if you need to see the picture big then

Serggg [28]

"the answer will be 0.00 "

3 0

3 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

That dissolve in water and has salty taste

Amiraneli [1.4K]

Answer:

it is salt an I right or not

6 0

3 years ago

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Which of the following would NOT diffuse through the plasma membrane by means of simple diffusion?1 oxygen2 glucose3 a steroid h

Mumz [18]

Answer:

Option 2= Glucose

Explanation:

Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.

Facilitated diffusion:

it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.

Primary active transport:

The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.

Secondary active transport:

It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.

5 0

3 years ago