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3 years ago

Plz help me thank you so much for helping

Chemistry

2 answers:

malfutka [58]3 years ago

6 0

The answer to it is A

Rina8888 [55]3 years ago

3 0

Answer:

A: matter or mass cannot be created or destroyed

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what is the percent yield of NaCl if 31.0 g of CuCl reacts with excess NaNo3 to produce 21.2 g of NaCl

olchik [2.2K]

Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.

Balanced Equation: CuCl + NaNO₃   → NaCl + CuNO₃

Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol

the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,

then moles of NaCl = 0.31 mol

Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g

⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.

Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %

NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.

4 0

3 years ago

If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

MakcuM [25]

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

5 0

3 years ago

34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i

nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

p(r) = $Ar(r^{2}e^{\frac{-r}{3a_{o}}}$

= $Ar^{3}e^{\frac{-r}{3a_{o}}}$

$\frac{dp(r)}{dr}$ = 0

$Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}}$ + $Ae^{\frac{-r}{3a_{o}}}(3r^{2})$ = 0

$\frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}$

= $3Ar^{2}e^{\frac{-r}{3a_{o}}}$

r = $9a_{o}$

Thus, we can conclude that most likely radius at which the electron would be found is $9a_{o}$ .

4 0

3 years ago

Please help me as soon as possible! Please help me as soon as possible!

Nastasia [14]

The Answer is D. Suspending a heavy weight with a strong chain.

6 0

3 years ago

Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric

kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

$pressure\propto \frac{1}{volume}$ (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0

3 years ago