How many atoms of hydrogen does alcohol (2C2H6O) have?

choli [55]

Answer:

$\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}$

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

$\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

Multiply by the given number of moles: 15.2

$15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

The moles of potassium cancel.

$15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }$

The denominator of 1 can be ignored.

$15.2 * {6.022*10^{23} \ atoms \ K }{$

Multiply.

$9.15344*10^{24} \ atoms \ K$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

$9.15*10^{24} \ atoms \ K$

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

3 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

4 years ago

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Olin [163]

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M

6 0

1 year ago

15g of FeCI3 is dissolved in 450 mL of solution. What is the concentration of [CI-]?

storchak [24]

The concentration of [CI-] : 0.617 M

<h3>Further explanation</h3>

FeCl₃ dissolved in 450 mL of solution(will dissociate )

Reaction

FeCl₃⇒Fe³⁺+3Cl⁻

mol FeCl₃(MW=162,2 g/mol)

$\tt \dfrac{15}{162.2}=0.0925$

mol Cl⁻ :

$\tt \dfrac{3}{1}\times 0.0925=0.2775$

molarity of Cl⁻ :

$\tt \dfrac{0.2775}{0.45}=0.617~M$

7 0

3 years ago

Dehydrohalogenation of 1-chloro-1-methylcyclopropane affords two alkenes (A and B) as products.

UNO [17]

Explanation:

Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:

Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.

Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.

However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.

Elimination of the hydrogen from the methyl group is easier.

Thus, the major product will A

4 0

3 years ago