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amm1812
3 years ago
6

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

7 0
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
8 0
3 years ago
In a similar experiment, a current of 2.15 amps ran for 8 minutes and 24 seconds. The temperature of the water was 26.0°C. The v
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}}}

3 0
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
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