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3 years ago

What is the mass (in grams) of 9.09x10^24 molecules of methanol (CH3OH)?

1 answer:

8 0

The amount of a substance can be converted into grams of that substance by using the molar mass. It is a physical property of a substance which describes the mass per mole of the substance. The molar mass of CH3OH is 32.04 g/ mol. Multiplying the given amount of CH3OH with the molar mass will yield to the mass of the substance.

<span>9.09x10^24 molecules CH3OH ( 1 mol / 6.022x10^23 molecules) (32.04 g/mol) = 483.63 g CH3OH

Dividing the molar mass of N2O which is 44.08 g/mol to the given amount will yield to the number of moles.

0.183 g N2O / 44.08 g/ mol =0.0042 mol N2O</span>

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Imagine that one tree outside your home looks unhealthy, although all the other trees seem healthy and strong. Describe how you

Westkost [7]

Answer:

Hypothesis---experiments----results----conclusion.

Explanation:

First we make a hypothesis means a statement about why the tree looks unhealthy. In this segment of scientific method we have to test the hypothesis through experimentation. After that we have to take the readings of various parts of the tree and analyze the data to find out the problem. In the next step, we have to made the results on the basis of the data that is obtained. In the last we have to write the conclusion of the analysis and see the hypothesis.

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3 years ago

A transverse wave is traveling from north to south. Which statement could be true for the motion of the wave particles in the me

Agata [3.3K]

<span>C. Their direction of motion is south and north. </span>

7 0

3 years ago

Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid w

Naddika [18.5K]

Answer:

87.9 % is the percent yield of H₂O

Explanation:

This is the neutralization reaction. A base reacts with an acid to produce water and the correspondly ionic salt.

NaOH + HCl → NaCl + H₂O

As we have the mass of the two reactants, we must determine the limiting reactant.

Let's convert to moles, the mass of each reactant. (mass / molar mass)

21.1 g / 36.45 g/mol = 0.579 moles of HCl

46.3 g / 40g/mol = 1.15 moles of NaOH

Ratio is 1:1, so it is obviously that the limiting reactant is the HCl. For 1.15 moles of NaOH, i need the same amount of acid, but I only have 0.579 moles

Let's work with the products now. Ratio is 1:1 again, so If I have 0.579 moles of acid, I can produce 0.579 moles of H₂O.

How many grams are 0.579 moles of water? We should find it out as this

mol . molar mass = mass → 0.579 mol . 18 g/mol = 10.4 g

We were told that the production of water was 9.17 g, so let's determine the percent yield as this:

(Yield produced / Theoretical yield) . 100 =

(9.17 g / 10.4g ) . 100 = 87.9 %

6 0

3 years ago

Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L

const2013 [10]

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

4 0

3 years ago

Read 2 more answers

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

3 years ago