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kakasveta [241]
3 years ago
12

Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.

Chemistry
1 answer:
velikii [3]3 years ago
6 0
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
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Which metal will react spontaneously with Cu2+ (aq) at 25°C?
Gwar [14]

Answer:

Mg  

Explanation:

The standard reduction potentials are

                                              <u>E°/V </u>

Au³⁺(aq ) + 3e⁻  ⟶  Au(s);     1.42

Hg²⁺(aq)  + 2e⁻  ⟶  Hg(l);     0.85

Ag⁺(aq)    +   e⁻  ⟶  Ag(s);    0.80

Cu²⁺(aq)   + 2e⁻ ⟶  Cu(s);   0.34

Mg2+(aq) + 2e- ⟶  Mg(s);   -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

 

5 0
3 years ago
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The following reaction is at equilibrium in a sealed container.
ale4655 [162]

Answer:

D.Lowering the temperature is the best option.

Explanation:

The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.

In the reaction below for example;

A + B <==>C+D

If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.

3 0
3 years ago
22.00 g of CO2 has a volume of 50.00 L and a pressure of 0.8210 atm. What must be the
creativ13 [48]

Answer:

T = 9.875K

Explanation:

The ideal gas Law is PV = nRT.

P = Pressure

V = Volume

n = amount of substance

R =  8.314 J/(K. mol)

T = Temperature in Kelvin

22g CO2

CO2 Molar Mass = 44g/mol

C = 12g/mol

O = 16g/mol

P = 0.8210atm

V = 50L

PV = nRT

0.8210 x 50 = 8.314 x 0.5 x T

41.05 =  4.157T

T = 41.05/4.157

T = 9.875K

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3 years ago
Explain how our body systems work together to get oxygen into and around our body. Answer must be three sentences long
sashaice [31]

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8 0
3 years ago
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What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
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Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

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3 years ago
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