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4 years ago

Identify the correct same regarding the strength of chemical bonds

Chemistry

1 answer:

bulgar [2K]4 years ago

7 0

Answer:

weak bonds require less energy to form bonds than strong bonds

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Your boss tells you that she needs a decision by the end of the day about the machine you want to purchase for your new operatio

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Option 1 : 1900 - 1786 = 114

Option 2 : 1450 - 1300 = 150

Option 3: 1950 - 1833 = 117

Option 4: 2000 - 1822 = 178

Option 1 has the lowest energy rate so ver time would be the best deal.

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4 years ago

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Answer:

1 is correct and 2 is liquid

Explanation:

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3 years ago

What does HHPS stand for?

maw [93]

Answer:

it stands for Hazardous Household Products Symbols.

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4 years ago

Which subatomic particles are found in the nucleus of an atom of beryllium?

timofeeve [1]

A. the nucleus carries a positive charge, and the electron cloud carries a negatuve charge. since neutrons carry no charge, the nucleus composed of neutrons and protons carry a positive charge.

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3 years ago

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Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water from –10 °C to 182°C. Assume that the specif

TiliK225 [7]

Answer : The amount of heat required is, $1.16\times 10^6J$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(182^oC)$

The expression used will be:

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat required for the reaction

m = mass of ice = 346 g

$c_{p,s}$ = specific heat of solid water or ice = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.184J/g^oC$

$c_{p,g}$ = specific heat of gaseous water = $1.99J/g^oC$

$\Delta H_{fusion}$ = enthalpy change for fusion = $333J/g$

$\Delta H_{vap}$ = enthalpy change for vaporization = $2260J/g$

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

$\Delta H=[346g\times 2.09J/g^oC\times (0-(-80))^oC]+346g\times 333J/g+[346g\times 4.184J/g^oC\times (100-0)^oC]+346g\times 2260J/g+[346g\times 1.99J/g^oC\times (180-100)^oC]$

$\Delta H=1.16\times 10^6J$

Therefore, the amount of heat required is, $1.16\times 10^6J$

4 0

3 years ago