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Andreas93 [3]
2 years ago
13

How much 1.5 m sodium hydroxide is necessary to exactly neutralize 20.0 ml of 2.5 m phosphoric acid?

Chemistry
1 answer:
andrew-mc [135]2 years ago
8 0
Answer is: 100 mL of <span>sodium hydroxide.
Chemical reaction: 3NaOH + H</span>₃PO₄ → Na₃PO₄ + 3H₂O.
V(H₃PO₄) = 20.0 mL ÷ 1000 mL/L = 0.02 L.
n(H₃PO₄) = V(H₃PO₄) · c(H₃PO₄).
n(H₃PO₄) = 0.02 L · 2.5 mol/L.
n(H₃PO₄) = 0.05 mol.
From chemical reaction: n(H₃PO₄) : n(NaOH) = 1 : 3.
n(NaOH) = 0.15 mol.
V(NaOH) = n(NaOH) ÷ c(NaOH).
V(NaOH) = 0.15 mol ÷ 1.5 mol/L.
V(NaOH) = 0.1 L · 1000 mL/L = 100 mL.
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Calculate the mass (in grams) of cobalt that has the same number of atoms as 521 mg of silicon.
denis23 [38]
The atomic mass of silicon is 28 g/mol.
Therefore, 521 mg or 0.521 g will be equivalent to;
  0.521/28 = 0.01861 mooles
Atomic mass of cobalt is 27 g/mol
According to Avogadro's law the mass of cobalt that will have the same atom as 52 mg of silicon will be ;
(0.01861 moles × 27) = 0.5024 g or 502.4 mg

8 0
2 years ago
If a pharmacist added 12 g of azelaic acid diluent should be used to prepare 8 fluidto 50 g of an ointment containing 15% ounces
natta225 [31]
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.



7 0
2 years ago
Select the correct answer. A student places a sample of white sodium bicarbonate powder in a test tube and heats it. The student
inessss [21]

Answer is: D. It is not sodium bicarbonate.

Balanced chemical reaction of heating sodium bicarbonate:                    2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.

This is chemical change (chemical reaction), because new substances are formed (sodium carbonate, carbon(IV) oxide and water), the atoms are rearranged, so there is no sodium bicarbonate (NaHCO₃) in the test tube.

8 0
2 years ago
Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. if 4.40 mol of m
Georgia [21]
The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2
8 0
2 years ago
2CH4(g)⟶C2H4(g)+2H2(g)
Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of CH_4 will be,

2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)    \Delta H^o=?

The intermediate balanced chemical reaction will be,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890.3kJ

(2) C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)     \Delta H_2=-136.3kJ

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=-571.6kJ

(4) 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)     \Delta H_4=-3120.8kJ

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) 2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)     \Delta H_1=2\times (-890.3kJ)=-1780.6kJ

(2) C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)    \Delta H_2=-(-136.3kJ)=136.3kJ

(3) H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ

(4) 2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)     \Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ

The expression for enthalpy of the reaction will be,

\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)

\Delta H=201.9kJ

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0
1 year ago
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