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jonny [76]
3 years ago
11

In a laboratory 0.500 moles of NAOH is completely dissolved in distilled water to form 400.0 mL of NAOH. The solution is then us

ed to titrate a solution of HNO3.
Which answer choice correctly completes the equation by filling in the formulas of the products in this titration reaction?

NaOH(aq) + HNO3(aq) —> —————- + —————-


1: NaNO2(aq) + H2O2(L)

2: NaNO3(aq) + H2O(L)

3: HNaNO3(aq) + OH(L)

4:H2NO3(aq) + NaO(L)
Chemistry
2 answers:
Eduardwww [97]3 years ago
5 0

Answer:

The answer to your question is number 2.  NaNO₃   +       H₂O

Explanation:

The reaction that is represented in this problem is an acid-base reaction. An acid-base reaction is a double displacement reaction.

The cation of a reactant attaches to the anion of the other reactant and viceversa.

Reactants      NaOH  and    HNO₃

Cations          Na⁺                 H⁺

Anions          OH⁻                NO₃⁻

Following the previous rule

Products       NaNO₃   +       H₂O

The products are a salt and water.

V125BC [204]3 years ago
4 0

Answer:

NaNO2(aq) + H2O2(l)

Explanation:

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A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

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8 0
3 years ago
You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the
Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

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8 0
3 years ago
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

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The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

3 0
3 years ago
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Goryan [66]

Answer:

6 neutrons

Explanation:

6 neutrons

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7 0
2 years ago
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uysha [10]

Answer:

Gravity

Explanation:

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5 0
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