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____ [38]
3 years ago
11

A solution was prepared by dissolving 31.0 g of kcl in 225 g of water. part a calculate the mass percent of kcl in the solution.

Chemistry
1 answer:
iragen [17]3 years ago
3 0
<span>Answer: Mass % KCL: Add the grams of both compounds (31.0 g KCL + 225 g water) to find total mass and then divide the grams of KCL over the total mass, then multiply by 100: ( 31.0 g KCL / 31.0 g + 225 G) * 100% Mole fraction KCL Calculate the moles of KCL and water and add them to find the total moles (Moles of KCL + moles of water). Then, divide the number of KCL moles over the total moles. moles of KCL/ moles kcl + moles water= mole fraction of KCL</span>
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What is the frequency and energy per quantum (in Joules) of :
viktelen [127]

(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

(b) f = 1.20 × 10¹⁰ Hz, E = 7.96 × 10⁻²⁴J.

<h3>Explanation</h3>

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at 3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1} in vacuum.

f = \dfrac{c}{\lambda}

where

  • f is the frequency of the electromagnetic radiation,
  • c is the speed of light, and
  • \lambda is the wavelength of the radiation.

(a)

Convert all units to standard ones.

\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}.

The unit of f shall also be standard.

f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}.

For each particle,

E = h\cdot f,

where

  • E is the energy of the particle,
  • h is the planck's constant where h = 6.63\times 10^{-34}\;\text{J}\cdot\text{s}^{-1}, and
  • f is the frequency of the particle.

E = h \cdot f = 6.63\times10^{-34}\;\text{J}\cdot\text{s}\times 5.00\times 10^{20}\;\text{s}^{-1} = 3.32\times10^{-13}\;\text{J}.

(b)

Try the steps in (a) for this beam of microwave with

  • \lambda = 2.50 \;\text{cm} = 2.50\times 10^{-2}\;\text{m}.

Expect the following results:

  • f = 1.20\times 10^{10}\;\text{Hz}, and
  • E = 7.96\times 10^{-24}\;\text{J}.
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4 0
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8 0
3 years ago
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How does the potential energy of a double bond (like oxygen-oxygen) compare to a weak interaction or single bond (like argon-oxy
astraxan [27]

Answer:

Potential energy in a double bond is going to be higher than the potential energy of a single bond due to the force needed to break a double bond as opposed to a single. Single bond is weaker, which means less energy is needed to break it. Double bonds are stronger, requiring more energy to break the bond.

Explanation:

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3 years ago
Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
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