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3 years ago

Which of the following best defines a mineral and a rock?

Chemistry

1 answer:

Dmitriy789 [7]3 years ago

3 0

Answer:

a. In a mineral the constituent atoms are bonded in a regular, repetitive, internal structure; a rock is a lithified or consolidated aggregate of different mineral grains.

Explanation:

In the internal structure of a mineral the atoms are arranged in an orderly manner, forming shapes that are repeated throughout the structure.

Rocks are a mixture of other materials consolidated in one, which include minerals.

So the minerals have a defined internal structure with atoms in a regular and repeated configuration, while a rock, being a consolidated aggregate, combination of other materials including minerals, does not have a structure that defined

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Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O).

Sindrei [870]

Oxen beacon is great for your self halllway

7 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

A 5.00-g sample of copper metal at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the copper is

vekshin1

Answer: The final temperature of the copper is 95°C.

Explanation:

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

$Q= m\times c\times \Delta T$

Q = heat absorbed = +133 J (heat is added to the system)

m = mass of copper = 5.00 g

c = specific heat capacity of copper = 0.38 J/g ° C

$\Delta T={\text{Change in temperature}}=T_2-T_1$

$T_1=25^oC$

Putting values in above equation, we get:

$+133J=5.00g\times 0.38J/g^oC\times (T_2-25)\\\\T_2=95^oC$

Hence, the final temperature of the copper is 95°C.

3 0

3 years ago

Which of the following combination of elements would result in covalent compound? * W X Y Z Vand X Wand Z Y and Z Wand y

TEA [102]

Answer:

C. Y & Z

Explanation:

V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).

Thus, Y and Z make covalent.

4 0

3 years ago

A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its

Over [174]

In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(0.1510 g) * 100
Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0

3 years ago