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3 years ago

Fault-block mountains are formed when a shift takes place along a fault line and

Chemistry

1 answer:

worty [1.4K]3 years ago

7 0

One side drops into fault

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What is one way to test whether and unknown solution is acidic or basic

dimulka [17.4K]

an acid is an ionic compound that produces positive hydrogen ions when dissolved in water and a bacis is an ionic compound that produces negative hydroxide ions when dissolved in water

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3 years ago

Read 2 more answers

Chemistry help please

Galina-37 [17]

Answer:

The reaction isn't yet at equilibrium. The overall reaction will continue to move in the direction of the products.

Assumption: this system is currently at $\rm 900^{\circ}C$ .

Explanation:

One way to tell whether a system is at its equilibrium is to compare its reaction quotient $Q$ with the equilibrium constant $K_c$ of the reaction.

The equation for $Q$ is quite similar to that for $K_c$ . The difference between the two is that $K_c$ requires equilibrium concentrations, while $Q$ can be calculated even when the system is on its way to equilibrium.

For this reaction,

$\displaystyle Q = \rm \frac{[CS_2]\cdot [H_2]^{4}}{[CH_4]\cdot [H_2S]^{2}}$ .

Given these concentrations,

$\displaystyle Q = \rm \frac{[CS_2]\cdot [H_2]^{4}}{[CH_4]\cdot [H_2S]^{2}} =\frac{1.51\times (1.08)^{4}}{1.15\times (1.20)^{2}} \approx 1.72$ .

The question states that at $\rm 900^{\circ}C$ , $K_c = 3.59$ . Assume that currently this system is also at $\rm 900^{\circ}C$ . (The two temperatures need to be the same since the value of $K_c$ depends on the temperature.)

It turns out that $Q = K_c$ . What does this mean?

First, the system isn't at equilibrium.
Second, if there's no external changes, the system will continue to move towards the equilibrium. Temperature might change. However, eventually $Q$ will be equal to $K_c$ , and the system will achieve equilibrium.

In which direction will the system move? At this moment, $Q < K_c$ . As time proceeds, the value of $Q$ will increase so that it could become equal to $K_c$ . Recall that $Q$ is fraction.

$\displaystyle Q = \rm \frac{[CS_2]\cdot [H_2]^{4}}{[CH_4]\cdot [H_2S]^{2}}$

When the value of $Q$ increases, either its numerator becomes larger or its denominator becomes smaller, or both will happen at the same time. However,

Concentrations on the numerator of $Q$ are those of the products;
Concentrations on the denominator of $Q$ are those of the reactants.

As time proceeds,

the concentration of the products will increase, while
the concentration of the reactants will decrease.

In other words, the equilibrium will move towards the products.

3 0

4 years ago

Need help ASAP!!!!!!!

geniusboy [140]

My Heroes is the time of my time I am in college now that I will do it again tomorrow night and you can do something to me about it and then I will never go back again I hope that I can do this again I don’t want

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3 years ago

lord [1]

Answer:

$\boxed{6.08 \times 10^{23}}$

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

T = 26.0 °C

V = 65.4 mL

p = 774.2 To

1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

$p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}$

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

$\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}$

3. Calculate the moles of electrons

$\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}$

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

$\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}$

6. Calculate Avogadro's number

$N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}$

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4 years ago

identify the postulate of Dalton's theory that is violated by the following observations: 59.95% of one sample of titanium dioxi

Inga [223]

<u>Answer:</u> This violates the law of constant composition.

<u>Explanation:</u>

Dalton's theory is based on mainly two laws, which are law of conservation of mass and law of constant composition.

Law of constant composition states that a compound always contain the elements in the fixed ratio by their masses.

For Example: In water $(H_2O)$ , the hydrogen and oxygen are present in the fixed ratio of 1 : 9 by their mass.

We are given:

A sample of titanium dioxide having 59.95 % of titanium and another sample of titanium dioxide having 60.10 % of titanium.

As, the compound is titanium dioxide. So, the mass percent of titanium must remain the same in both the elements.

Hence, this violates the law of constant composition.

8 0

4 years ago