The student's experiment most likely demonstrates how... A. earthquakes occur along a fault
B. new crusts move at midocean ridges
C. tectonic movements cause hotspot volcanism
D. mountains are created along cracks in Earth plates

Answers

Answer 1

Answer: The right answer for the question that is being asked and shown above is that: "D. mountains are created along cracks in Earth plates."The student's experiment most likely demonstrates how D. mountains are created along cracks in Earth plates

Answer 2

Answer:

Answer:d

Explanation:test

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Write the balanced equation for the reaction given below: C2H6 + O2 --> CO2 + H2O. If 16.4 L of C2H6 reacts with 0.980 mol of O2, how many liters of carbon dioxide gas will be produced? What is the limiting reagent? How many oxygen atoms will be in this sample of carbon dioxide? How many moles of the excess reactant will be left over? How many grams of the excess reactant will be left over?

Answers

The balanced reaction: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O

We first convert volume of C2H6 to no. of moles. We use the conditions at STP where 1 mol = 22.4 L thus,

Moles C2H6 = 16.4 L/ 22.4 L =0.7321 mol

In order to determine the limiting reagent, we look at the given amounts of the reactants.

0.7321 mol C2H6 (7/2 mol O2 / 1 mol C2H6) = 2.562 mol O2

From the given amounts of the reactants, we can say that O2 is the limiting reactant since we need 2.562 mol O2 to completely react the given amount of C2H6. The excess reagent is C2H6

To calculate for the amount of products and excess reactants:

0.980 mol O2 (2 mol CO2 / (7/2 mol O2)) = 0.56 mol CO2 (22.4 L / 1 mol ) =12.544 L CO2
0.980 mol O2 (1 mol C2H6 / (7/2 mol O2)) = 0.28 mol C2H6
Excess C2H6 = 0.7321 mol - 0.28 mol C2H6 = 0.4521 mol C2H6

We then use the molecular weight of C2H6 to convert the excess amount to grams.

0.4521 mol C2H6 (30.08 g C2H6 / 1 mol C2H6) = 13.60 g C2H6

Since the limiting reagent is O2 there will be no oxygen atoms that will be left after the reaction.

A typical neon light contains neon gas mixed with argon gas. If the total pressure of the mixture is 1.63 atm and the partial pressure of the neon is 0.71
atm. What is the partial pressure (atm) of argon?

Answers

the partial pressure of argon is:

1.63-0.71= 0.92 atm

1.63 atm - 0.71 atm = 0.92 atm

Which of the following is one way to prevent the corrosion of iron

Answers

D) its the easiest and most effective.

Which statement correctly describes the charge of the nucleus and the charge of the electron cloud of an atom?(1) The nucleus is positive and the electron cloud is positive.
(2) The nucleus is positive and the electron cloud is negative.
(3) The nucleus is negative and the electron cloud is positive.
(4) The nucleus is negative and the electron cloud is negative.

Answers

Atoms are made of nucleus and electrons. Nucleus have a positive charge due to the presence of protons and electrons are negative in charge . Thus option 2 is correct.

What is an atom?

Atoms are the basic unit every substance. Atoms combines to form molecules and compounds. An atom is made of nucleus and surrounding electrons which are revolving through a circular path of fixed energy.

Nucleus is the core of the atom and the 90% of the weight of atom is centered in the nucleus and it contains positively charged protons and neutral particles neutrons.

The nuclear model of atom was first proposed by Rutherford. He found the positive cloud of the atom and and named as nucleus. He discovered the positively charged particles protons.

The net charge of an atom is zero because all the positive charges are neutralised by equal number of negatively charged particles electrons.

Therefore, the nucleus is positive and the electron cloud is negative and option 2 is correct.

To find more about atomic nucleus, refer the link below:

brainly.com/question/10658589

#SPJ2

The nucleus of an atom is positive, because it contains the protons (+1 charge) and neutrons (0 charge)
The electron cloud has a negative charge, because electrons have a -1 charge
So the answer is (2)

Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2*10^-4

Answers

This problem uses the relationship between Kb and the the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:

Kb = KaKb
1.00 x 10^-14 = 7.2 x 10^-4(x)
x = 1.39 x 10^-11

We now need to calculate the [OH¯] using the Kb expression:

1.39 x 10^-11 = x^2 / (0.30 - x)

The denominator can be neglected. Thus, x is 3.73 x 10^-6.

pOH = -log 3.73 x 10^-6 = 5.43
pH = 14-5.43 = 8.57

Answer:

pH=8.32

Explanation:

The relevant equilibrium for this problem is

F⁻ + H₂O ↔ HF + OH⁻

With a constant Kb of

Kb= $([HF][OH^(-)])/([F^(-)])$

Kb= $(x*x)/(0.30-x)$

To calculate the value of Kb we use the formula Kw=Ka*Kb, where Kw is the ionization constant of water, 1 * 10⁻¹⁴.

1 * 10⁻¹⁴ = 7.2*10⁻⁴ * Kb

Kb = 1.4 * 10⁻¹¹

So now we have

1.4 * 10⁻¹¹= $(x*x)/(0.30-x)$

We make the assumption that x<<<0.30 M, so we can rewrite the equation of Kb as:

1.4 * 10⁻¹¹= $(x*x)/(0.30)$

$4.2*10^(-12)=x^(2) \nx=2.05*10^(-6)$

So [OH⁻]=2.05*10⁻⁶

pOH=5.68

pH = 14 - pOH

pH=8.32

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 1.50 g of butane?

Answers

Answer:

6.21 × 10²² Carbon Atoms

Solution:

Data Given:

Mass of Butane (C₄H₁₀) = 1.50 g

M.Mass of Butane = 58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 1.50 g ÷ 58.1 g.mol⁻¹

Moles = 0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

Moles = Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

Number of C₄H₁₀ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

Number of C₄H₁₀ Molecules = 0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

Number of C₄H₁₀ Molecules = 1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

1 Molecule of C₄H₁₀ contains = 4 Atoms of Carbon

So,

1.55 × 10²² C₄H₁₀ Molecules will contain = X Atoms of Carbon

Solving for X,

X = (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X = 6.21 × 10²² Atoms of Carbon

$\boxed{6.216 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ atoms}}}$ of carbon is present in 1.50 g of butane.

Further Explanation:

Avogadro’s number indicates how many atoms or molecules a mole can have in it. In other words, it provides information about the number of units that are present in one mole of the substance. It is numerically equal to ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$ . These units can be atoms or molecules.

The formula to calculate the moles of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ is as follows:

${\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \frac{{{\text{Given mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}{{{\text{Molar mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}$ …… (1)

Substitute 1.50 g for the given mass and 58.1 g/mol for the molar mass of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ in equation (1).

$\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} &= \left( {{\text{1}}{\text{.50 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.1 g}}}}} \right)\n&= {\text{0}}{\text{.0258 mol}}\n\end{aligned}$

Since one mole of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ has ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ . Therefore the formula to calculate the molecules of butane is as follows:

${\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \left( {{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}} \right)\left( {{\text{Avogadro's Number}}} \right)$ …… (2)

Substitute 0.0258 mol for the moles of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ and ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ for Avogadro’s number in equation (2).

$\begin{aligned}{\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\mathbf{}}&=\left( {0.0258{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}} \right)\n&= 1.554 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}} \n\end{aligned}$

The chemical formula of butane is ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ . This indicates one molecule of butane has four atoms of carbon. Therefore the number of carbon atoms can be calculated as follows:

$\begin{aligned}{\text{Atoms of carbon}} &= \left( {1.554 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}} \right)\left( {\frac{{{\text{4 C atoms}}}}{{{\text{1 molecule of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}} \right)\n&= 6.216 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ C atoms}} \n\end{aligned}$

Learn more:

Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603
Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Chapter: Mole concept

Subject: Chemistry

Keywords: 1.50 g, 58.1 g/mol, butane, C4H10, Avogadro’s number, $6.216*10^22$ C atoms, $1.554*10^22$ molecules, moles, one mole, chemical formula, carbon atoms, molar mass of C4H10, given mass of C4H10.