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2 years ago

What happens when baking soda and vinegar are mixed together.

Chemistry

1 answer:

egoroff_w [7]2 years ago

8 0

Answer: Mixing vinegar and baking soda initiates a chemical reaction that produces carbon dioxide and water.

Explanation:

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Write the electron configuration and Noble gas configuration for each element.

cestrela7 [59]

Answer:

1is the answer to that one day F

7 0

3 years ago

Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) -7.3

Andreyy89

Answer:

The pump work is 3451 kJ/s

Explanation:

Pump work (W) is calculated as

$W = (h_f - h_i) \times \dot{m}$

where

$h_f$ is the enthalpy of water at its final state

$h_i$ is the enthalpy of water at its initial state

$\dot{m} = 10 kg/s$ is water mass flow

For liquids, properties are evaluated as saturated liquid. From the figure attached, it can be seen that

$h_f = 762.81 kJ/kg$

$h_i = 417.46 kJ/kg$

Replacing

$W = (762.81 kJ/kg - 417.46 kJ/kg) \times 10 kg/s$

$W = 3451 kJ/s$

8 0

3 years ago

The following reaction was carried out in a 2.50 LL reaction vessel at 1100 KK: C(s)+H2O(g)⇌CO(g)+H2(g)C(s)+H2O(g)⇌CO(g)+H2(g) I

Schach [20]

Answer:

Q = 0.144

Explanation:

C(s) + H2O(g) ↔ CO(g) + H2(g)

reaction quotient:

Q = [H2(g)][CO(g)] / [H2O(g)][C(s)]

∴ [C(s)] = 10.0 mol/2.50 L = 4 M

∴ [H2O(g)] = 16.0 mol/2.50 L = 6.4 M

∴ [CO(g)] = 3.30 mol/2.50 L = 1.32 M

∴ [H2(g)] = 7.00 mol/2.50 L = 2.8 M

⇒ Q = (2.8 M)(1.32 M) / (6.4 M)(4 M)

⇒ Q = 0.144

3 0

4 years ago

What volume is occupied by 9.50 g c6h12 at stp (standard temperature and pressure)?

WINSTONCH [101]

Molar mass

6(12)+12(1)
7(12)
84g/mol

No of moles

Given mass/Molar mass
9.5/84
0.113mol

Volume

22.4L(0.113)
2.5L

4 0

2 years ago

A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$ $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$ $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

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

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

4 0

3 years ago