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3 years ago

What element is being oxidized in the following redox reaction?

Chemistry

1 answer:

gregori [183]3 years ago

4 0

Answer:

C is the element thats has been oxidized.

Explanation:

MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)

This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.

In the oxalic acid, this are the oxidation states:

H: +1

C: +3

O: -2

In the product side, in CO₂ the oxidation states are:

C: +4

O: -2

Carbon from the oxalate has increased the oxidation state, so it has been oxidized.

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A sample of potassium has an average atomic mass of 39.0983amu. There are three isotopic forms of potassium element in the sampl

igomit [66]

Answer:

Percent Composition of 41K = 6.7302%

Explanation:

The explination is in the image.

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3 years ago

R is the midpoint of FG.<br><br> FR = 26 find RG

vichka [17]

Answer:

Explanation:

This is because if R is the midpoint of FRG, FR is half of FRG, so basically all you do it multiply by 2 to get the FRG

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3 years ago

¿Qué ocurre cuando la radiación del sol atraviesa la atmósfera?

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3 years ago

The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.

Kitty [74]

Answer:

a) The theoretical yield is 408.45g of $BaSO_{4}$

b) Percent yield = $\frac{realyield}{408.45g}*100$

Explanation:

1. First determine the numer of moles of $BaCl_{2}$ and $Na_{2}SO_{4}$ .

Molarity is expressed as:

M= $\frac{molessolute}{Lsolution}$

- For the $BaCl_{2}$

M= $\frac{1.75molesBaCl_{2}}{1Lsolution}$

Therefore there are 1.75 moles of $BaCl_{2}$

- For the $Na_{2}SO_{4}$

M= $\frac{2.0moles[tex]Na_{2}SO_{4}$ }{1Lsolution}[/tex]

Therefore there are 2.0 moles of $Na_{2}SO_{4}$

2. Write the balanced chemical equation for the synthesis of the barium white pigment, $BaSO_{4}$ :

$BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl$

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the $BaCl_{2}$ :

$\frac{1.75}{1}=1.75$

- For the $Na_{2}SO_{4}$ :

$\frac{2.0}{1}=2.0$

As the $BaCl_{2}$ is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

$1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}$

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = $\frac{realyield}{theoreticalyield}*100$

Percent yield = $\frac{realyield}{408.45g}*100$

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0

3 years ago

The pressure of a sample of helium in a 200. ml container is 2.0 atm. If the helium is compressed to a pressure of 40. atm witho

Crank

Answer:

$V_2=10mL$

Explanation:

Hello there!

In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:

$P_2V_2=P_1V_1$

In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:

$V_2=\frac{P_1V_1}{P_2} \\\\V_2=\frac{2.0atm*200.mL}{40.atm}\\\\V_2=10mL$

Regards!

5 0

3 years ago