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kari74 [83]
3 years ago
11

Dan bikes 10 km west and then bikes another 5 km west. What is Dan's

Chemistry
1 answer:
777dan777 [17]3 years ago
4 0
C

This is because 10+5=15
15/45=0.3
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The specific heat capacity of liquid ethanol is 2.42 J/gºC.
VladimirAG [237]
Hope it helps for you : )

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3 years ago
What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same t
kenny6666 [7]

Answer: 20.7 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system=-P\Delta V  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = 4.51\times 4.184kJ=18.9kJ    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

\Delta E=+1.79+(18.9)=20.7kJ

Thus \Delta E for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ

4 0
3 years ago
What is the independent variable?<br> (For science)
DiKsa [7]

Answer:

The independent variable is what you change in the experiment.

Explanation:

The independent determines the dependent variable.

6 0
3 years ago
Human use of groundwater has _______ over time.
Harman [31]
Increased is the answer
7 0
3 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

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

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
4 years ago
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