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mylen [45]
3 years ago
15

Determine the hydroxide ion concentration in a solution that is 0.018 m naoh.

Chemistry
1 answer:
nydimaria [60]3 years ago
5 0
As we know, NaOH is a strong Base and completely dissociates in water as given below,

                            NaOH + H₂O    →   Na⁺ ₍aq₎  +  OH⁻ ₍aq₎

So, the concentration of OH⁻ Ions will be the same as that the concentration of NaOH i.e. 0.018 M.
So,
                                     [OH⁻]  =  0.018 M
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5. An example of a muscle that is voluntarily controlled is a muscle that
Shtirlitz [24]

Answer:

A, Makes the leg move

Explanation:

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3 years ago
If you had an object with a mass of 76g and a volume of 22 ml, what is its density? If you cut this object in half, what would b
V125BC [204]
D = M/V = 76g / 22ml = 3.4g/ml

Half ~ D = 38g / 11ml = 3.4g/ml

Even if the object you had was cut in half, it’s density would remain the same.
8 0
3 years ago
What mass of magnesium chloride is needed to make 100.0 mL of a solution that is 0.500 M in chloride ion?
miss Akunina [59]
M = n/V

.5M = n/.100 L

n = .1 L * .5M

n= .05 mols of MgCl2

mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2 

mass of MgCl2 = 4.76 grams

4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused
3 0
3 years ago
How much heat is required to change temperature of 10 g of water from 4 °C to 8 °C? (Water has a specific heat of 4.18 )?
Andrews [41]

Answer: 167.2 J

Explanation:

5 0
2 years ago
2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo
pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

<u>Chemical reaction involved</u>: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K    (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol)     (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2

<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}

\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)

\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)

<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>

4 0
3 years ago
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