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lina2011 [118]
2 years ago
15

What would be the resulting molarity of a solution made by dissolving 45.7 grams of Mg(OH)2 in enough water to make a 1087-milli

liter solution? Show all of the work needed to solve this problem
Chemistry
2 answers:
Tom [10]2 years ago
7 0

<u>Answer:</u> The molarity of the solution is 0.72 M

<u>Explanation:</u>

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of magnesium hydroxide = 45.7 g

Molar mass of magnesium hydroxide = 58.32 g/mole

Volume of solution = 1087 mL

Putting values in above equation, we get:

\text{Molarity of the solution}=\frac{45.7\times 1000}{58.32\times 1087}=0.72M

Hence, the molarity of the solution is 0.72 M

liq [111]2 years ago
3 0
Molar mass Mg(OH)2 = <span>58.3197 g/mol

Number of moles:

45.7 / 58.3197 => 0.7836 moles

Volume = 1087 / 1000 => 1.087 L

Therefore:

Molarity = moles / volume

Molarity = </span>0.7836 / 1.087

=> 0.7208 M

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Write the equations that represent the first and second ionization steps for hydroselenic acid (H2Se) in water. (Use H3O+ instea
Otrada [13]

Answer:

The equations are

1) H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

2) HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

Explanation:

There are two ionization steps in the dissociation of hydroselenic acid.

In first dissociation the H₂Se loses one proton and forms hydrogen selenide ion as shown below:

H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

The next step is again removal of a proton from the base formed above.

HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

4 0
2 years ago
analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula
MrMuchimi
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr   
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr  /0.0135= 1 mol Cr
 0.0538 mol O/0.035= 4 mol Cr
K2CrO4
6 0
2 years ago
Read 2 more answers
The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG
Dmitrij [34]

Answer:

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP -----------> ADP + creatine phosphate    

ΔG∘   = -12.6 KJ/mole  = -12600 J/mole

ΔG = -0.1 KJ/mole  =  -100 J/mole

[Creatine phosphate]  = 25 mM = 25*10^-3 M

[Creatine] = 17 mM    = 17*10^-3 M

[ATP]   =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600   = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500   = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc   = 6.44*10^-3

But,

Qc    = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3   = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

          = 2.1896 *10^-5 M

          = 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

3 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c
makkiz [27]
0.216g of aluminium compound X  react with an excess of water water to produce gas. this gas burn completely  in O2  to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n  =  24 / 1 
n = 0.0045 mole ( CO2 >>0.0045 mole 
0.216 - 0.0045 = 0.2115
so Al =   0.2115 / 27  =>  0.0078 mole 
C = 0.0045 * 1000 => 4.5    and Al  = 0.0078 * 1000 = 7.8 

7 0
2 years ago
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