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-BARSIC- [3]
3 years ago
12

1. Write the balanced chemical equation for the synthesis of magnesium phosphide from its elements, including the word “energy”

as a product.
- Write the Keq expression for the reaction.
Chemistry
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

3Mg(s) +2P(s) -------> Mg3P2(s) + energy

Keq= [Mg3P2]/[Mg]^3 [P]^2

Explanation:

The equation for the formation of magnesium phosphide from its elements is;

3Mg(s) +2P(s) -------> Mg3P2(s) + energy

Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.

The equilibrium expression for the reaction K(eq) will be given by;

Keq= [Mg3P2]/[Mg]^3 [P]^2

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nydimaria [60]

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<h3>What are the effects of summer in plants in Michigan and Florida?</h3>

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5 0
2 years ago
A green powder was placed into a beaker of water. After it was stirred
bekas [8.4K]

Answer:

A solution was NOT formed.

Explanation:

For a solution to be formed, the green powder would need to dissolve in the water to form a homogenous mixture. But since the water was cloudy and had lumps of the green powder, it means that the powder did not dissolve, but instead remained insoluble.

7 0
3 years ago
Consider the reaction 3X + 2Y 5C + 4D
xeze [42]

Answer:

This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the molesof X that are needed to 

produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:Mole ratios:In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.

Calculations:In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:

Explanation:

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5 0
2 years ago
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
3 years ago
HELP!!!! ASAP!!!!!!!!!!<br><br> HOW ARE REACTANTS DIFFERENT FROM THE PRODUCTS ?
raketka [301]
Reactants are what make up the product.
7 0
3 years ago
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