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3 years ago

Calculate the Density of the substances below to determine if they will float or sink in water.

Chemistry

1 answer:

Reil [10]3 years ago

7 0

Explanation:

If the density of the object is more than that of water, it will sink. Otherwise it will float. The density of water is 1 g/mL.

Substance 1,

Mass, m = 450 g, Volume, V = 90 mL

Density = mass/volume

So,

$d_1=\dfrac{450}{90}\\\\=5\ g/mL$

It will sink.

Substance 2,

Mass, m = 35 g, Volume, V = 70 mL

Density = mass/volume

So,

$d_2=\dfrac{35}{70}\\\\=0.5\ g/mL$

It will float.

Substance 3,

Mass, m = 24 g, Volume, V = 12 mL

Density = mass/volume

So,

$d_3=\dfrac{24}{12}\\\\=2\ g/mL$

It will sink.

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Lars is balancing equations with his study group. He is unsure about one equation because each member of the study group came up

Stells [14]

Answer:

3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O

Explanation:

This a proper way to balance the equation:

- Count the OH from the base (2)

- Count the H from the acid (3)

We can make 2 molecules of H₂O but we still have one more H

H₃PO₄ → dissociates in → 3H⁺ + PO₄³⁻

Ca(OH)₂ → dissociates in → Ca²⁺ + 2OH⁻

So, to form the salt, you must have 3 Ca²⁺ to react with 2 (PO₄³⁻) to make global charge of +6/-6

Therefore, if you have 3 Ca in the salt, you may have 3 Ca in the base.

So, if you have 2 phosphate in the salt, you must have 2 PO₄³⁻ in the acid.

Now you have 6 protons in the acid (6H) and 6 (OH) in the base; in conclussion you can make 6 H₂O.

Finally the ballance equation is:

3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O

8 0

4 years ago

Read 2 more answers

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

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

3 0

3 years ago

Write the equilibrium constant expression, Kc, for the following reaction: Please enter the compounds in the order given in the

xeze [42]

Answer:

See the answer below, please.

Explanation:

The equilibrium constant is defined as the relationship between products and reagents, each one elevated to their stoichiometric coefficients, in that of the given equation, the Kc is:

Kc= (NH4)^1/ (NH3)^1 x (HI)^1

NH4= products

NH3 and HI = reagents

7 0

3 years ago

If a person swallows acetone what are the procedures that should be followed

sergij07 [2.7K]

Answer:

get poison control

Explanation:

3 0

3 years ago

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The atomic masses of any two elements contain the same number of

den301095 [7]

Answer:

They contain of atoms

Explanation:

That's because atomic weights or masses of each atom of each element are proportional to each other, the same number of atoms of each element will give masses that are also proportional to each other. If you start with 20 oxygen atoms, you will need 40 hydrogen atoms to make the water and you will get 20 molecules of water.

8 0

3 years ago