All categories

2 years ago

How does the amount of energy coming from the energy source (low vs. high) affect the amount of electrical energy created.

1 answer:

6 0

Answer:The environmental problems directly related to energy production and Electricity is a secondary energy source that is generated producedfrom primary energy sources:

hope that helps im new so have a good day (:

Explanation:

You might be interested in

Help, 8th grade Science

valentinak56 [21]

Answer:

Control container

Explanation:

5 0

2 years ago

Read 2 more answers

What can be concluded from the following statements about a property of metats? • Acast-iron skillet on a stove is used to try b

zlopas [31]

Answer:

D - metals conduct heat

7 0

2 years ago

analysis of a 12.04g sample of liquid compound compoed of carbon,hydrogen and nitrogen showed it to contain 7.34g carbon ,1.85g

SVEN [57.7K]

Answer:

Nskalakababiakviaug

4 0

3 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

2 years ago

One of the hydrates of MnSO4 is manganese(II) sulfate tetrahydrate . A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in

Dennis_Churaev [7]

Answer:

48.32 g of anhydrous MnSO4.

Explanation:

Equation of dehydration reaction:

MnSO4 •4H2O --> MnSO4 + 4H2O

Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)

= 223 g/mol

Mass of MnSO4 • 4H2O = 71.6 g

Number of moles = mass/molar mass

= 71.6/223

= 0.32 mol.

By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4

Number of moles of MnSO4 = 0.32 mol.

Molar mass = 55 + 32 + (4*16)

= 151 g/mol.

Mass = 151 * 0.32

= 48.32 g of anhydrous MnSO4.

3 0

2 years ago