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3 years ago

What is the percentage yield if 125.4g C3H8 are collected from a reaction that should produce 157.4g C3h8?

Chemistry

1 answer:

cestrela7 [59]3 years ago

3 0

The percentage yield : 79.67%

The closest answer is option B.

<h3>Further explanation</h3>

Given

125.4 g C3H8

157.4 g C3H8

Required

The percentage yield

Solution

Percent yield is the comparison of the amount of product obtained from a reaction with the amount you calculated

General formula:

% yield = (actual yield : theoretical yield) x 100%

Input the value :

% yield = (125.4 : 157.4) x 100%

% yield = 79.67

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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l

Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law: $PV=nRT$ where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units: $R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means $T=273.15K$ and $P=1atm$

Lastly, we must calculate the number of moles of $O_2(g)$ there are. Given 0.80g of $O_2(g)$ , we will need to convert with the molar mass of $O_2(g)$ . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: $32g\text{ }O_2=1mol\text{ }O_2$

Thus, $\frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n$

Isolating V in the Ideal Gas Law:

$PV=nRT$

$V=\frac{nRT}{P}$

...substituting the known values, and simplifying...

$V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}$

$V=0.56L \text{ } O_2$

So, 0.80g of $O_2(g)$ would occupy 0.56L at STP.

5 0

2 years ago

Read 2 more answers

What is the formula for number of moles?

kicyunya [14]

Answer:

number of moles=volume/molar volume or mass/molar mass

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3 years ago

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The maximum number of electrons in a single d subshell is:

Iteru [2.4K]

10 electrons

Explanation:

The maximum number of electrons in a single d-subshell is 10 electrons.

The d-sub-orbital used to denote azimuthal or secondary quantum numbers.

The maximum number of electrons in the orbitals of sublevels are:

two electrons in the s-sublevel, it has one orbital

six electrons in the p-sublevel, it has three orbital

ten electrons in the d- sublevel, it has five orbitals

fourteen electrons in the f-sublevel, it has seven orbitals

The maximum number of electrons in an orbital is two.

learn more:

Atomic orbitals brainly.com/question/1832385

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3 years ago

Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

Thus, for all the given combinations, we obtain:

- Y⁻

$X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3$

- Y²⁻

$X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3$

- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

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8 0

3 years ago

This question in the screenshot

Morgarella [4.7K]

Mass of hydrate + crucible = 47.29 g

Mass of anhydrous salt = 2.7 g

Molar mass of anhydrous salt CuSO4 = 159.5 g

Given,

mass of empty crucible = 42.45 g

mass of hydrate salt= 4.84 g

mass of crucible after first heating = 46.1 g

mass of crucible after second heating= 45.153 g

mass of crucible after third heating= 45.15 g

so, as per the question we need to find...

Mass of hydrate + crucible = ? g

Mass of anhydrous salt = ? g

Molar mass of anhydrous salt CuSO4 = ? g

∴Mass of hydrate + crucible = 42.45 + 4.48 = 47.29 g

The given salt is in hydrate form, to remove water from this molecule we need to perform heating .

So we are taking the substance into the crucible as it is in less quantity.

Here, we performed heating 3 times and note the weight after every heating.

After this, assume that the water is totally evaporated and the remaining salt is in anhydrous form,

∴ Mass of anhydrous salt = 45.15 - 42.45 = 2.7 g

To find the molar mass of anhydrous salt of CuSO4,

atomic weight of Cu = 63.5 g

atomic weight of S = 32 g

atomic weight of O =16 g

∴ molar mass of anhydrous salt of CuSO4 = 63.5 + 32 + (16 ×3)

=159.5 gm

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5 0

2 years ago