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3 years ago

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Bromine is is reddish-brown liquid that boils at 59 degrees. bromine is highly reactive with many metals, for example, it reacts

with sodium to form a white solid. classify each of these properties of bromine as either a physical or a chemical property

1 answer:

katrin2010 [14]3 years ago

4 0

A physical property of an element is a property of an element that can observed or measured without changing the chemical nature of the element.

A chemical property of an element is a property of an element that can only be observed or measure when the chemical property of the element is altered or changed.

Based on this;
The boiling point of bromine is a physical property of bromine.
The high reactivity of bromine with many elements is a chemical property of bromine.

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Which is a negative effect of deforestation on the environment? habitat restoration, desertification, soil production, or urbani

skad [1K]

Answer:

The correct option is desertification

Explanation:

Deforestation is the deliberate act of pulling down trees to provide a clear land for space or to make use of the trees for other purposes such as grazing, farming, timber fire wood, building and constructions, urban utilization, access roads, expansion, development of infrastructures, grazing of cattle and mining

The effects of deforestation includes climate change, flooding, increase in the proportion of green house gases in the atmosphere, soil erosion and desertification.

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3 years ago

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KonstantinChe [14]

Answer:

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3 years ago

Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t

Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of $HC_2H_3O_2 \ (M_1)$ = 0.105 M

Volume of $HC_2H_3O_2 \ (V_1)$ = 20.0 mL

Concentration of $NaOH (M_2)$ = 0.125 M

The chemical reaction can be expressed as:

$HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}$

Using the ICE Table to determine the equilibrium concentrations.

$HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}$

I 0.105 0 0

C -x +x +x

E 0.105 - x x x

$K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}$

$K_a = \dfrac{(x)(x)}{(0.105-x)}$

Recall that the ka for $HC_2H_3O_2= 1.8 \times 10^{-5}$

Then;

$1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}$

$1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}$

By solving the above mathematical expression;

x = 0.00137 M

$H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )$

pH = 2.86

Hence, the initial pH = 2.86

b) To determine the volume of the added base needed to reach the equivalence point by using the formula:

$M_1 V_1 = M_2 V_2$

$V_2= \dfrac{M_1V_1}{M_2}$

$V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}$

$V_2 = 16.8 mL$

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of $HC_2H_3O_2 =$ molarity × volume

$= 0.105 \ M \times 20.0 \times 10^{-3} \ L$

$= 2.1 \times 10^{-3}$

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = $0.625 \times 10^{-3}$

After reacting with 5.0 mL NaOH, the number of moles is as follows:

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Initial moles $2.1*10^{-3}$ $0.625 * 10^{-3}$ 0 0

F(moles) $(2.1*10^{-3} - 0.625 \times 10^{-3})$ 0 $0.625 \times 10^{-3}$ $0.625 \times 10^{-3}$

The pH of the solution is then calculated as follows:

$pH = pKa + log \dfrac{[base]} {[acid]}$

Recall that:

pKa for $HC_2H_3O_2=4.74$

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

$pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}$

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the $(C_2H_3O^-_2)$ with $H_2O$

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of $(C_2H_3O^-_2)$ = moles/volume

= $\dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}$

= 0.0571 M

Now, using the ICE table to determine the concentration of $H_3O^+$ ;

$C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}$

I 0.0571 0 0

C -x +x +x

E 0.0571 - x x x

Recall that the Ka for $HC_2H_3O_2$ = $1.8 \times 10^{-5}$

$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} } \\ \\ K_b = 5.6 \times 10^{-10}$

$k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}$

$5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}$

$x = [OH^-] = 5.6 \times 10^{-6} \ M$

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }$

$[H_3O^+] =1.77 \times 10^{-9}$

$pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }$

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Before 0.0021 0.002725 0

After 0 0.000625 0.0021

$[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}$

$[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}$

$[OH^-] = 0.0149 \ M$

From above; we can determine the concentration of $H_3O^+$ by using the following method:

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}$

$[H_3O^+] = 6.7 \times 10^{-13}$

$pH = - log [H_3O^+]$

$pH = -log (6.7 \times 10^{-13} )$

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

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3 years ago

Which law relates temperature pressure volume and the number of moles of a gas under perfect conditions?

Zepler [3.9K]

I think the correct answer would be Ideal gas law. The law that relates temperature pressure volume and the number of moles of a gas under perfect conditions. It <span>is the equation of state of a hypothetical </span>ideal gas<span>. It is a good approximation of the behavior of many </span>gases<span> under many conditions. Hope this answers the question.</span>

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4 years ago

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aivan3 [116]

Answer:

I think is D. The same atoms must be in the products and the reactants.

Explanation:

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