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tigry1 [53]
3 years ago
10

Temperature of 50g of methanol from 20 degrees Celsius to 70 degrees Celsius

Chemistry
1 answer:
olganol [36]3 years ago
4 0

[ Q ] = [ Delta U ] + [ W sub B ]

No work is done. -----> [ W sub B ] = 0.0

[ Delta U ] = [ m ] [ C sub V ] [ T2 - T1 ]

[ Delta U ] = [ 50 ] [ 0.58 ][70 - 20 ]

[ Delta U ] = 58 cal

Q = [ Delta U ] = 58cal
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***URGENT PLEASE**** Select all of the bases listed *
Masteriza [31]

Answer:

NaOH (a strong base)

Ammonia ( weak base )

6 0
3 years ago
If your end product is 1.5 moles of KMnO 4 how many moles of manganese oxide were used in the reaction? The equation for the pro
vovikov84 [41]

Answer:

1.5 moles

Explanation:

The equation of the reaction is given as:

2 MnO2 + 4 KOH + O2 --> 2KMnO 4 + 2KOH + H2

From the equation,

2 moles of MnO2 produces 2 moles of KMnO4

x moles of MnO2 would produce 1.5 moles of KMnO4

2 = 2

x = 1.5

Solving for x;

x = 1.5 * 2 / 2

x = 1.5 moles

4 0
3 years ago
4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a
yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0
3 years ago
Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this
maw [93]

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol

Thus the enthalpy change for the given reaction is -705.55kJ

7 0
3 years ago
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