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Margaret [11]
3 years ago
8

How many grams of glucose, C6H12O6, in 2.47 mole?

Chemistry
2 answers:
konstantin123 [22]3 years ago
7 0
Answer:
number of grams = 444.6 grams

Explanation:
From the periodic table, we can find that:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
This means that:
molar mass of C6H12O6 = 6(12) + 12(1) + 6(16) = 180 grams

Now, number of moles can be calculated using the following rule:
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass = 2.47 * 180
mass = 444.6 grams

Hope this helps :)
statuscvo [17]3 years ago
6 0
First, we need to find the atomic mass of C_{6}H_{12}O_{6}.

According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16

As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of  C_{6}H_{12}O_{6} = 6 * 12.01 + 12 * 1.008 + 6 * 16

The molar mass of  C_{6}H_{12}O_{6} = 180.156 grams/mole

Now that we have the molar mass of  C_{6}H_{12}O_{6}, we can find the grams of glucose by using:

mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)

Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams

Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams

-i
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The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x
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planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
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3 years ago
If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

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What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?
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The system is isothermal, so we use the formula:
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Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ 
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