All categories

3 years ago

How are silicate minerals different from nonsilicate minerals

1 answer:

7 0

Answer: The main difference between silicate minerals and nonsilicate minerals is that silicate minerals are composed of silicate groups whereas Nonsilicate minerals have no silicate groups.

Explanation:

You might be interested in

1. Which of the following best describes the relationship

Ipatiy [6.2K]

D there is one kind of cell of which all living things are made

8 0

3 years ago

My teacher is grading this soon can someone help me ASAP!

Stolb23 [73]

10. You demonstrated the difference in density of the two objects. It is a physical property.

11. First calculate the density for all of them: density = mass/volume

Density:

A. 5/6 g/ml

B. 10/9 g/ml

C. 15/16 g/ml

D. 20/10 g/ml

If the density of the substance is higher than the density of the substance it is put in, then it will sink. So substances B and D will sink in water, as their densities are higher than 1 g/ml.

12. Ammonia weighs less than water does-- for example, the weight of 8 gallons of ammonia will be equivalent to the weight of 5 gallons of water.

Hope this helped!

3 0

3 years ago

Liquid hexane

maks197457 [2]

Answer: The mass of $H_2O$ produced is 2.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

For hexane:

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

$\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol$

For oxygen gas:

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol$

The chemical equation for the combustion of hexane follows:

$2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O$

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = $\frac{19}{2}\times 0.020=0.19mol$ of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of $H_2O$

So, 0.020 moles of hexane will produce = $\frac{14}{2}\times 0.020=0.14mol$ of $H_2O$

We know, molar mass of $H_2O$ = 18 g/mol

Putting values in above equation, we get:

$\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g$

Hence, the mass of $H_2O$ produced is 2.52 g

4 0

3 years ago

How many neutrinos does an element have if it's atomic number is 47 and it's mass number is 169

Stells [14]

This element would have 122 neutrons.

Atomic mass is the accumulation of neutrons and protons.

169 - 47 = 122.

Therefore, there are 122 neutrons.

7 0

3 years ago

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

V = 0.750 liter
M = 0.480 mol/liter
M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 liter
M = 0.290 mol/liter
M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0

3 years ago