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3 years ago

Please Help Weak Acid Base Problems!!!

Chemistry

2 answers:

TiliK225 [7]3 years ago

8 0

A it is a______________________

DiKsa [7]3 years ago

3 0

Answer:

The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the sample problem. However, the variable x will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.

Explanation:

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How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? CH4(

Vedmedyk [2.9K]

Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₄: 1 mole
O₂: 2 moles
CO₂: 1 mole
H₂O: 2 moles

The molar mass of the compounds is:

CH₄: 16 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₄: 1 mole ×16 g/mole= 16 grams
O₂: 2 moles ×32 g/mole= 64 grams
CO₂: 1 mole ×44 g/mole= 44 grams
H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

$mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}$

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

$mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}$

<u><em>mass of CO₂= 10.93 grams</em></u>

Then, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

Learn more about the reaction stoichiometry:

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6 0

2 years ago

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

Korvikt [17]

Answer:

$55.2kgNa_{3}AlF_{6}$

Explanation:

1. First balance the equation for the synthesis of cryolite:

$Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}$

2. Find the limiting reagent between the $Al_{2}O_{3},NaOH$ and $HF$

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}$

$55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH$ $55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF$

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

$Al_{2}O_{3}:\frac{131}{1}=131$

$NaOH:\frac{1385}{6}=231$

$HF:\frac{2770}{12}=231$

The $Al_{2}O_{3}$ is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}$

3 0

4 years ago

A plate and frame filter press with 12 frames is used to filter slurry at constant pressure. Total filtration area is 65 ft2 and

Vladimir [108]

Answer:

$t_2 = 14.4 min$

Explanation:

Given data:

Number of frame is 12+ 8 = 20

Total filtrate is 360 lb

we know

For 1 plate area $= \frac{65}{12} ft^2$

Then for 20 are PLATE $= 20\tmes \frac{65}{12} ft^2 = 108.33 ft^2$

filtrate equation is given as

$t\frac{\mu_s\times(\alpha v) V^2}{A^2(-\Delta P)}$

In the above equation only t and A is variable , all other terms are constant

Therefore we have

$\frac{t_2}{t_1}= \frac{A_1}{A_2} = [\frac{65}{105.33}]^2 = 0.36$

$t_2 = 0.36\times 40 = 14.4 min$

$t_2 = 14.4 min$

5 0

3 years ago

Decide which element probably has a boiling point most and least similar to the boiling point of cesium.

natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.

<em>Hope that helps!</em>

5 0

3 years ago

A solution that may contain Cu2+, Bi3+, Sn4+, or Sb3+ ions is treated with thioacetamide in an acid medium. The black precipitat

Ede4ka [16]

Answer:

Sb3+ is present.

Ions that could be absent include copper II ions(Cu2+) and lead II ions(Pb2+)

The procedure for finding ions in doubt is explained below under explanation.

Explanation:

Sb3+ is the only ion that's definitely present. This group 2B sulfide is the only one that will be orange in colour when re - precipitated from the hydroxide ion(OH-) extract. Nevertheless, the orange color could serve as a masking of the yellow Arsenic trisulphide and the Tin sulphide precipitates, but not the black mercury sulphide precipitate.

Now, back to the black original precipitate. For it to be so, a group IIA cation must be present. The only group IIB cation with a black sulfide is mercury cation Hg2+ which we said could not be masked earlier.

The group IIA cations that must be present are Pb2+, Cu2+, Cd2+, Bi3+. When the acid extract of any of these 4 cations are present are made alkaline with NH3 then Copper ion (Cu2+) and Cadmium ion (Cd+) will form soluble amine complexes, which is very possible.

Bismuth Cation(Bi3+) would form the white precipitate of Bi(OH)3 and Lead II cation(Pb2+). However, if they are not removed in Group I, they will form solid Pb(OH)2, which is also white.

To find out the ions that are still in doubts. For example, when adding Ammonia(NH3) noted earlier to the Group IA extract, if the solution turns blue, then copper II ion(Cu2+) is confirmed otherwise, it is absent or present in very low concentration. However, if we first treated the precipitate with Sodium Hydroxide(NaOH), and it dissolves, then it confirms that Pb2+. Bi(OH)3 will not dissolve in Sodium Hydroxide(NaOH) solution.

8 0

3 years ago