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3 years ago

Is this right if not pls lmk what the correct answer is:)

Chemistry

1 answer:

trapecia [35]3 years ago

5 0

Answer:

It's right

Explanation:

Hopefully this helps!

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Does gravity exist without air?

sveticcg [70]

Mass is required for gravity. Higher mass implies higher gravitational drag. Most of these thoughts are tied to other planets without atmospheres, or the moon. The moon doesn't have any air or atmosphere and it pulls the Earth water.

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3 0

3 years ago

The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 101.3 k

Gnom [1K]

The H2 produced in a chemical reaction is collected through water in a eudiometer. The pressure (kPa) of the H2 gas is 98.89 kPa

The total pressure in a chemical reaction is the total sum of the partial pressure and the vapor pressure of the chemical substances taking place in the chemical reaction.

Total pressure = partial pressure of H₂ gas + vapor pressure of H₂O

∴

The vapor pressure of H₂ gas = Total pressure in the eudiometer - partial pressure of H₂O

Given that:

The total pressure in the eudiometer = 101.3 kPa
The partial pressure of H₂O = 2.41 kPa

The vapor pressure of H₂ gas = 101.3 kPa - 2.41 kPa

The vapor pressure of H₂ gas = (101.3 - 2.41) kPa

The vapor pressure of H₂ = 98.89 kPa

Therefore, we can conclude that the vapor pressure of H₂ is 98.89 kPa.

Learn more about partial pressure here:

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2 years ago

16. Which element has th17. Magnesium, Mg, tends to form a +2 ion. Which will have the larger atomic radius? Mg or Mg2+ (Explain

SashulF [63]

Answer:

Th is the symbol for element Thorium.

Mg is magnesium while Mg 2+ is magnesium Ion. Judging from periodic trends, atomic radius is one half the distance between the atoms of two covalently bonded atoms. It decreases as elevtrons are added to valence shells. That means, across the period it increases and down the group it decreases. Making Mg2+ smaller.

Electronegativity is the energy needed to take an electron in the gaseous state. Florine is more electron negative that lithium. It increases across the period and decreases down the group. Except in the Noble group.

Explanation:

6 0

4 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

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

What is the mole ratio of iron (III) chloride (FeCl3) to zinc chloride (ZnCl2) ?

tatiyna

Answer:C

Explanation:

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3 years ago