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larisa [96]
2 years ago
7

Convert molecules of vitamin d (molar mass 384.7 g/mol) to grams of vitamin d.

Chemistry
1 answer:
velikii [3]2 years ago
3 0

The 6.023 * 10²³  molecule means  384.7 grams of Vitamin D.

<h3>What is a mole ?</h3>

A mole is a universal unit used to measure large number of atoms , molecules of an element.

one mole is equal to the Avogadro's Constant.

Mole is equal to the ratio of the mass of a compound to its molecular weight

It is given that Vitamin D is present

no. of molecule = 6.023 * 10²³

6.023 * 10²³ molecule constitutes of 1 mol

1 mole  is equal to the molar mass of a compound/ element

Therefore 1 mole = 384.7grams of Vitamin D

and hence 6.023 * 10²³  molecule means 384.7 grams of Vitamin D.

To know more about Mole

brainly.com/question/26416088

#SPJ1

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At STP, fluorine is a gas and iodine is a solid. This observation can be explained by the fact that fluorine has
motikmotik
<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
6 0
3 years ago
Read 2 more answers
Kc for the reaction N2O4 &lt;=&gt; 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0
3 years ago
2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)
Paha777 [63]
<h3>Answer:</h3>

16.7 g H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                               \displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})
  2. Multiply/Divide:                 \displaystyle 16.6685 \ g \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

16.6685 g H₂O ≈ 16.7 g H₂O

6 0
3 years ago
How many moles are in a 111.5 gram sample of sodium chloride, NaCl?
Katarina [22]

Answer: 1.91 moles

Explanation:

First you want to find the molar mass of NaCL

Na = 22.99g  Cl = 35.45g

22.99g + 35.45g = 54.44g

Then divide 111.5g by 54.44g and this will give you moles.

5 0
3 years ago
Find the volume in milliliters of 2.00 mol of an ideal gas at 36°C and a pressure of 1120 torr.
hram777 [196]

Answer:

V = 34430 mL

Explanation:

Given data:

Volume in mL = ?

Number of moles of gas = 2.00 mol

Temperature = 36°C (36+273= 309K)

Pressure of gas = 1120 torr

Solution:

Formula:

PV = nRT

V = nRT/P

V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr

V = 38563.2 torr • L / 1120 torr

V = 34.43 L

L to mL

34.43 L ×1000 mL / 1 L

34430 mL

5 0
3 years ago
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