Hi, i need help.. please

amm1812

For example, iron plus oxygen can become ferric oxide. Both elements change their names. The change is used to indicate the kind of bonding process that is taking place. When iron and oxygen become ferric oxide, the iron has lost electrons and the oxygen has gained the electrons that iron lost.

4 0

3 years ago

Read 2 more answers

A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

5 0

3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago

After a polypeptide chain has been synthesized, certain amino acids in the peptide may become modified. For each modified amino

AnnZ [28]

Answer:

The modification that yields the first amino acid depicted is lysine

Explanation:

Lysine being an amino acid used in preparation of various medicine. It is used to prevent cold as well as sore throat, which is found to be caused by the herpes simplex labialis. It can be used orally or can be directly applied over the skin. various foods have lysine in it and they are meat, cheese, certain fishes, eggs, soybeans, spirulina and also fenugreek seed. They are the building block for the protein particles.

5 0

3 years ago

Read 2 more answers

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

6 0

3 years ago