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vivado [14]
2 years ago
14

Apply Concepts H2O and H2O2 are binary molecular compounds generally known by their common names, “water” and “hydrogen peroxide

.” Following the naming conventions you identified for molecular compounds, what would their names be? Explain your reasoning.
Chemistry
1 answer:
labwork [276]2 years ago
4 0

Answer:The two-party system came into being because the structure of U.S. elections, with one seat tied to a geographic district, tends to lead to dominance by two major political parties. ... Plurality voting has been justified as the simplest and most cost-effective ... After the election is over, supporters experience remorse when

Explanation: hope u get it right

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what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require
vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of HCOO^{-}

So, moles of NaOH used to reach equivalence point equal to number of moles HCOO^{-} produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of HCOO^{-} produced = \frac{29.80\times 0.3567}{1000}moles=0.01063moles

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of HCOO^{-} = \frac{0.01063\times 1000}{54.80}M=0.1940M

At equivalence point, pH depends upon hydrolysis of HCOO^{-}. So, we have to construct an ICE table.

HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, \frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}

species inside third bracket represent equilibrium concentrations

So, \frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}

or,x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0

So, x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}

So, x=3.285\times 10^{-6}M

So, pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52

5 0
3 years ago
Which option describes the behavior of energy in exothermic reactions?
jok3333 [9.3K]

Answer:

The energy of the products is less than the energy of the reactants, so energy is released into the surrounding environment.

6 0
3 years ago
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Olegator [25]

Answer:

I know that the 100-mL graduated cylinders are always read to 1 decimal place.

I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.

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