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BartSMP [9]
2 years ago
8

A 103.8g sample of nitric acid solution that is 70.0% HNO3 contains by mass

Chemistry
1 answer:
soldi70 [24.7K]2 years ago
3 0
w/w percentage <span>
               = mass of the pure compound / total mass of the sample x 100%

70% HNO₃ contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>

The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
                                         = 103.8 g x (70 / 100)
<span>                                         = 72.66 g = 72.7 g.</span></span>
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Which of the following statements is true about energy quantization at the atomic level? Electrons in the outermost orbits are t
MAVERICK [17]
<h2>Answer:</h2>

The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".

<h3>Explanation:</h3>
  • There are different orbitals around the nucleus on which the electrons moves around the nucleus.
  • These orbitals have a specific energy, due to which they are known as energy levels.
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A chunk of dry ice, solid CO2, "disappears" after sitting at room temperature for a while. There is no puddle of liquid. What ha
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I believe the answer is A.
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2 years ago
A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
Llana [10]

<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
                                    = </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

4 0
2 years ago
Which of the following statements is true? (Multiple answers possible) Group of answer choices:
Sophie [7]

Answer:The endpoint does not correspond exactly to the equivalence point

At the endpoint, a change in a physical quantity associated with the equivalence point occurs.

At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present.

Explanation:

The end point is always indicated by some physical property that changes such as colour. At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present. The equivalence point cannot be physically observed but can be deduced after a titration curve is plotted.

3 0
2 years ago
3) Calculate the percent by mass of 3.55 g NaCl dissolved in 88 g water.
kotykmax [81]

Answer:

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

Explanation:

When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.

A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:

Percent by mass=\frac{mass of solute}{mass of solution} *100

In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.

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Replacing:

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Percent by mass= 3.88%

<u><em>The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%</em></u>

5 0
2 years ago
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