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Ulleksa [173]
3 years ago
12

Based on symmetry, which bond is most likely to be nonpolar in nature?

Chemistry
1 answer:
oee [108]3 years ago
8 0
B be cause all of their electronegativity cancel on another and no matter how you fold it, it will still be the same
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Which of the following statements about the Moon is true?
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Calculate the number of moles 3.16 g of KMnO4 32.33 g of K2CrO4 100 g of KHCO3
kvasek [131]
The number of moles:
3.16g of KMnO_{4}
1 mole of KMnO_{4}- 158g.
x moles of KMnO_{4}- 3.16g.
x=\frac{1\ mole*3.16g}{158g}=\frac{3.16}{158}=0.02\ mole

The number of moles:
32.33g of K_{2}CrO_{4}
1 mole of K_{2}CrO_{4}- 194g.
x moles of K_{2}CrO_{4}- 32.33g.
x=\frac{1\ mole*32.33g}{194g}=\frac{32.33}{194}=0.17\ mole

The number of moles:
100g of KHCO_{3}
1 mole of KHCO_{3}- 100.1g.
x moles of KHCO_{3}- 100g.
x=\frac{1\ mole*100g}{100.1g}=\frac{100}{100.1}=0.999\ mole


Greetings, 
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3 0
2 years ago
1.00 kg of ice at -10 °C is heated using a Bunsen burner flame until all the ice melts and the temperature reaches 95 °C. A) How
BaLLatris [955]

Answer : The energy required is, 574.2055 KJ

Solution :

The conversions involved in this process 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)(95^oC)

Now we have to calculate the enthalpy change or energy.

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

where,

\Delta H = energy required = ?

m = mass of ice = 1 kg  = 1000 g

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

c_{p,l} = specific heat of liquid water = 4.18J/g^oC

n = number of moles of ice = \frac{\text{Mass of ice}}{\text{Molar mass of ice}}=\frac{1000g}{18g/mole}=55.55mole

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

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

\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC]

\Delta H=574205.5J=574.2055kJ     (1 KJ = 1000 J)

Therefore, the energy required is, 574.2055 KJ

3 0
3 years ago
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