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3 years ago

Why there is not reaction in nac2h3o2(aq)+pb(no3)2(aq)→?

1 answer:

4 0

NaC2H3O2 is called sodium acetate (CH3COONa)

2CH3COONa(aq) + Pb(NO3)2(aq) -> 2NaNO3(aq) + (CH3COO)2Pb (aq)

The reaction will not occur because no products will be precipitated, like one of the reactants is in solid form

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In a lead-acid battery, the electrodes are consumed. In this battery, Select one: a. the anode is PbO2. b. the cathode is Pb. c.

Gnoma [55]

Answer:

The correct answer is option e.

Explanation:

The lead acid battery consists lead as an anode and lead oxide as a cathode. Both the electrodes are suspended in dilute sulfuric acid which act as an electrolyte.

At anode: Oxidation

$Pb(s)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2e^-$

At cathode: Reduction

$PbO_2+SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow PbSO_4(s)+2H_2O$

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3 years ago

How many neutrons in beryllium

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Answer:

There are 5 neutrons in beryllium

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3 years ago

The following reaction occurs in aqueous solution: NH4 + (aq) + NO2 - â†’ N2 (g) + 2H2O (l) The data below is obtained at 25Â°C.

GREYUIT [131]

Answer: The order with respect to $NH_4^+$ is 1.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$NH_4^++NO_2^-\rightarrow N_2+2H_2O$

$Rate=k[NH_4^+]^x[NO_2^-]^y$

k= rate constant

x = order with respect to $NH_4^+$

y = order with respect to A $NO_2^-$

n = x+y = Total order

From trial 1: $3.2\times 10^{-3}=k[0.0100]^x[0.200]^y$ (1)

From trial 2: $6.4\times 10^{-3}=k[0.0200]^x[0.200]^y$ (2)

Dividing 2 by 1 : $\frac{6.4\times 10^{-3}}{3.2\times 10^{-3}}=\frac{k[0.0100]^x[0.2000]^y}{k[0.0200]^x[0.200]^y}$

$2=2^x,2^1=2^x$ therefore x= 1

Thus order with respect to $NH_4^+$ is 1.

8 0

3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

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3 years ago