When the Kelvin temperature of an enclosed gas doubles, the particles of the gas

Question 2 (1 point)

babunello [35]

Answer:

o and cl

Explanation:

metals and nonmetals tend to form ionic bonds

7 0

3 years ago

What is the ionic equation for the following 2FeCl(aq) 3Mg(s) 3MgCl(aq) 2Fe(s)

amid [387]

Answer : The net ionic equation will be,

$2Fe^{3+}(aq)+3Mg(s)\rightarrow 3Mg^{2+}(aq)+2Fe(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$2FeCl_3(aq)+3Mg(s)\rightarrow 3MgCl_2(aq)+2Fe(s)$

The ionic equation in separated aqueous solution will be,

$2Fe^{3+}(aq)+6Cl^{-}(aq)+3Mg(s)\rightarrow 3Mg^{2+}(aq)+6Cl^{-}(aq)+2Fe(s)$

In this equation, $Cl^{-}$ is the spectator ion.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Fe^{3+}(aq)+3Mg(s)\rightarrow 3Mg^{2+}(aq)+2Fe(s)$

7 0

3 years ago

For the reaction: CO(g) + H2O(g) â‡Œ CO2(g) + H2(g) the value of Kc is 1.845 at a specific temperature. We place 0.500mol CO and

patriot [66]

Answer:

Explanation:

In the equilibrium:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

Kc is:

Kc = 1.845 = [CO₂] [H₂] / [CO] [H₂O]

Where [] are equilibrium concentrations of each species

Initial concentrations:

[CO] = 0.500mol / 1.00L = 0.500M

[H₂O] = 0.500mol / 1.00L = 0.500M

In equilibrium, concentrations will be:

[CO] = 0.500M - X

[H₂O] = 0.500M - X

[CO₂] = X

[H₂] = X

Where X is reaction coordinate. The amount of reactant that reacts producing products.

Replacing in Kc expression:

1.845 = [CO₂] [H₂] / [CO] [H₂O]

1.845 = [X] [X] / [0.500M - X] [0.500M - X]

1.845 = X² / X² - X + 0.25

1.845X² - 1.845X + 0.46125 = X²

0.845X² - 1.845X + 0.46125 = 0

Solving for X:

X = 0.288M; Right solution

X = 1.9M; False solution: Produce negative concentrations.

Replacing, equilibrium concentrations are:

[CO] = 0.500M - X = 0.212M

[H₂O] = 0.500M - X = 0.212M

[CO₂] = X = 0.288M

[H₂] = X = 0.288M

3 0

3 years ago

Plsss help ASAP Dont guess

alexira [117]

Answer:

B

Explanation:

5 0

3 years ago

Read 2 more answers

When 20.00 mL of an unknown monoprotic acid is titrated with 0.125 M NaOH, it takes 15.00 mL to reach the endpoint. What is the

ohaa [14]

Answer:

About 0.0940 M.

Explanation:

Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

$\displaystyle \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)} \longrightarrow \text{H$_2$O}_\text{($\ell$)} + \text{A}^-_\text{(aq)}$

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

$\displaystyle \begin{aligned} 15.00\text{ mL} &\cdot \frac{0.125\text{ mol NaOH}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} \\ \\ &\cdot \frac{1\text{ mol OH}^-}{1\text{ mol NaOH}} \cdot \frac{1\text{ mol HA}}{1\text{ mol OH}^-}\\ \\ & = 0.00188\text{ mol HA}\end{aligned}$

Therefore, the molarity of the original solution was:

$\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}$

In conclusion, the molarity of the unknown acid is about 0.0940 M.

3 0

3 years ago