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3 years ago

If 11.87 grams of sand (SiO2) were contained in the mixture, how many atoms of oxygen were in the mixture?

Chemistry

1 answer:

ozzi3 years ago

6 0

Answer:

2.38x10²³ atoms oxygen

Explanation:

To solve this question we need to convert the mass of sand to moles using its molar mass (Molar mass SiO₂ = 60.08g/mol). Twice these moles are the moles of Oxygen and using Avogadro's number we can find the amount of atoms of Oxygen in the mixture:

Moles SiO₂:

11.87g * (1mol / 60.08g) = 0.1976moles SiO₂

Moles Oxygen:

0.1976moles SiO₂* 2 = 0.3951moles oxygen

Atoms oxygen:

0.3951moles oxygen * (6.022x10²³atoms / 1mol O₂) =

<h3>2.38x10²³ atoms oxygen</h3>

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The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

3 years ago

How does the density of a gas depend on the molar mass of the gas?

AlladinOne [14]

Answer:

The density of the ideal gas is directly proportional to its molar mass.

Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

$\rho = \frac{m}{V}$ ......equation (1)

According to the ideal gas law for ideal gas:

$PV = nRT$ ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

The number of moles: $n = \frac{m}{M}$

where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

$PV = \frac{m}{M}RT$

⇒ $PM= \frac{m}{V} RT$

⇒ $\frac{PM}{RT}= \frac{m}{V}$ ......equation (3)

Now from equation (1) and (3), we get

$\frac{PM}{RT}= \frac{m}{V} = \rho$

⇒ Density of an ideal gas: $\rho = \frac{PM}{RT}$

⇒ Density of an ideal gas: ρ ∝ molar mass of gas: M

Therefore, the density of the ideal gas is directly proportional to its molar mass. 

6 0

3 years ago

A baseball has a mass of 135 grams and a softball has a mass of 270 grams. In which of the following situations would they have

liraira [26]

Answer:

The baseball is thrown twice as fast as the softball in the same direction.

Explanation:

8 0

3 years ago

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be

mylen [45]

yield = 52.23 %

Explanation:

We have the following chemical reaction:

2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)

mass of aluminium = mass of bottle with aluminium pieces - bottle mass

mass of aluminium = 10.8955 - 9.8981 = 0.9974 g

mass of alum = mass of bottle with final product - bottle mass

mass of alum = 19.0414 - 9.8981 = 9.1433 g

number of moles = mass / molecular weight

number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles

To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if 2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 / 0.03694) × 100

yield = 52.23 %

Learn more about:

reaction yield

brainly.com/question/7786567

#learnwithBrainly

5 0

3 years ago

What is oxidized in a galvanic cell with aluminum and gold electrodes ?

k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

$Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V$

$Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive): $Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Notice that the overall cell potential upon summing is:

$E_{cell}=1.66 V + 1.50 V=3.16 V$

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0

3 years ago