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xz_007 [3.2K]
3 years ago
12

PLEASE HELP ASAP MOLES TO MOLECULES

Chemistry
2 answers:
Vlada [557]3 years ago
7 0

Answer:

4.77mol is the correct answer

kupik [55]3 years ago
5 0
4.77 mol, is the answer:)
You might be interested in
Determining Density and Using Density to Determine Volume or Mass
Shalnov [3]

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answer:

density = \frac{mass}{volume}

ρ=m/v ,m=ρv,    v=m/ρ

(a)m=1*10g  , v=7.36cm³

    ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

   v= 65/0.791 =82.17g/mL

(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³

    m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

   ρ=374.5/41.8 =8.96g/cm³

 mass of ethanol=15g,  density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

4 0
3 years ago
A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

8 0
2 years ago
Find the mass in grams of 4.52
Tanzania [10]

Answer:

Mass = 1274 .64 g it would be option C if it is converted into kilogram

1274 .64 / 1000 = 1.27 Kg

Explanation:

Given data:

Number of moles of C₂₀H₄₂ = 4.52 mol

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1.0 g/mol

Mass of C₂₀H₄₂ = ?

Solution:

Number of moles = mass / molar mass

Molar mass = 20× 12 + 42× 1.0 = 282 g/mol

Now we will put the values in formula:

Number of moles = mass / molar mass

4.52 mol = mass / 282 g /mol

Mass = 4.52 mol × 282 g/mol

Mass = 1274 .64 g

7 0
3 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
Answer the question below based on the periodic table entry for bromine.
Shalnov [3]
<span>The number of neutrons bromine will have are equal to
= protons + neutrons
so,
80-35=45</span>
4 0
3 years ago
Read 2 more answers
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