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lorasvet [3.4K]

2 years ago

11

If the pOH of vinegar is 9.45, what is its [OH− ]?

2 answers:

7nadin3 [17]2 years ago

4 0

Answer:
[OH⁻] = 3.54 × 10⁻¹⁰ M

Solution:

pOH is related to [OH⁻] as,

pOH = - log [OH⁻]

Putting value of pOH,

9.45 = -log [OH⁻]

Solving for [OH⁻],

[OH⁻] = 10⁻⁹·⁴⁵ ∴ 10 = Antilog

[OH⁻] = 3.54 × 10⁻¹⁰ M

Rainbow [258]2 years ago

3 0

Answer:

3.55 and -10

Explanation:

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Chlorine + sulfur dioxide + water = hydrochloric acid + sulfuric acid

scZoUnD [109]

The given example is a chemical reaction.

The contents (separated as reactants and products) :

$\begin{tabular}{c | l}Reactants & Products \\\cline{1-2}Chlorine & Sulfuric Acid \\Water & Hydrochloric Acid \\Sulfur Dioxide & \\\end{tabular}$

The written reaction is :

$\boxed {Cl + SO_{2} + H_{2}O \implies HCl + H_{2}SO{4}}$

<em>I hope it helped you solve the problem.</em>

<em>Good luck on your studies!</em>

7 0

1 year ago

Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h

Gwar [14]

<u>Answer:</u>

<em>0.264 g of $CO_2$ can be formed from 288 mg of $O_2$ </em>

<u>Explanation:</u>

The balanced chemical equation is

$2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O$

The conversions are

Mass in mg $O_2$ is converted to mass in g $O_2$

Mass in g $O_2$ is converted to moles $O_2$ by dividing with molar mass

Moles $O_2$ is converted to moles $CO_2$ by using the mole ratio of $O_2:CO_2$ is 9 : 6

Moles $CO_2$ is converted to mass $CO_2$ by multiplying with molar mass $CO_2$

mass in mg $O_2$ > mass in g $O_2$ >moles $O_2$ > moles $CO_2$ > mass $CO_2$

$288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}$

=0.264g (Answer)

7 0

3 years ago

What is the unknown metal if the temperature of a beaker of 100ml of water was raised 17c to 19 c when 21 grams of the metal at

horrorfan [7]

Answer:

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

Explanation:

<u>Step 1</u>: Data given

Mass of the metal = 21 grams

Volume of water = 100 mL

⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams

Initial temperature of metal = 122.5 °C

Initial temperature of water = 17°C

Final temperature of water and the metal = 19 °C

Heat capacity of water = 4.184 J/g°C

<u />

<u>Step 2: </u>Calculate the specific heat capacity

Heat lost by the metal = heat won by water

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)

21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)

-2173.5 *c(metal) = -836.8

c(metal) = 0.385 J/g°C

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

4 0

3 years ago

Which of the following environments is most likely to have days with a relative humidity of less than 10% (ten percent)?

Anon25 [30]

Answer:

Desert

Explanation:

Desert barely have any water therefore they can barely produce any amount of humidity

6 0

2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago