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3 years ago

Help me with an easy question worth 25 points

Chemistry

2 answers:

swat323 years ago

7 0

Answer: Velocity

Explanation:

The density of a material affects the speed that a wave will be transmitted through it. In general, the denser the transparent material, the more slowly light travels through it. Glass is denser than air, so a light ray passing from air into glass slows down.

Blababa [14]3 years ago

7 0

Answer:

C is most dense i think lol

Explanation:

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Can you find a salmander in the desert or rainforest?

vodka [1.7K]

You can find it in rainforest because they like damp places.

6 0

3 years ago

7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat

Pie

Answer:

11.4g of S₂Cl₂ is the expected yield

9.69g of S₂Cl₂ are produced with a 85% yield

Explanation:

The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:

S₈ + 4Cl₂ → 4S₂Cl₂

<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>

To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:

S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈

Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂

For a complete reaction of 0.0390 moles of sulfur, there are necessaries:

0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.

As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:

0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =

<h3>11.4g of S₂Cl₂ is the expected yield</h3>

If you produce just the 85.0% of yield, mass of S₂Cl₂ is:

11.4g ₓ 85% =

3 0

3 years ago

▸ Exam Instructions

neonofarm [45]

C. They’re safe to use without the regulation of a physician, according to a regulatory agency

7 0

2 years ago

1.674×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.6 s. Under identical conditions, 1.715×10-

siniylev [52]

<h2>Answer:</h2>

44.06 g/mol

<h3>Explanation:</h3>

We are given;

Number of moles of unidentified gas as 1.674×10^-4 mol
Time of effusion of unidentified gas 86.6 s
Number of moles of Argon gas as 1.715×10^-4 mol
Time of effusion of Argon gas is 84.5 s

We are supposed to calculate the molar mass of unidentified gas

<h3>Step 1: Calculate the effusion rates of each gas</h3>

Effusion rate = Number of moles/time

Effusion rate of unidentified gas (R₁)

= 1.674×10^-4 mol ÷ 86.6 s

= 1.933 × 10^-6 mol/s

Effusion rate of Argon gas (R₂)

= 1.715×10^-4 mol ÷ 84.5 sec

= 2.030 × 10^-6 mol/s

<h3>Step 2: Calculate the molar mass of unidentified gas</h3>

Assuming the molar mass of unidentified gas is x;
We can use the Graham's law of effusion to find x;
According to Graham's law of diffusion;

$\frac{R_{1}}{R_{2}}}=\frac{\sqrt{MM_{Ar}}}{\sqrt{X}}$

But, Molar mass of Argon is 39.948 g/mol

Therefore;

$\frac{1.933*10^-6mol/s}{2.030*10^-6mol/s}}=\frac{\sqrt{39.948}}{\sqrt{X}}$

$0.9522=\frac{\sqrt{39.948}}{\sqrt{X}}$

Solving for X

x = 44.06 g/mol

Therefore, the molar mass of the identified gas is 44.06 g/mol

3 0

4 years ago

Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

kumpel [21]

Answer : The limiting reactant is $O_2$

Explanation : Given,

Mass of $C_3H_8$ = 30.0 g

Mass of $O_2$ = 75.0 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_3H_8$ and $O_2$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction we conclude that

As, 5 mole of $O_2$ react with 1 mole of $C_3H_8$

So, 2.34 moles of $O_2$ react with $\frac{2.34}{5}\times 1=0.468$ moles of $C_3H_8$

From this we conclude that, $C_3H_8$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Therefore, the limiting reactant is $O_2$

5 0

3 years ago