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Choose the correct mole ratio for the balanced equation below.<br><br> Ca + 2H2O → Ca(OH)2 + H2

Cerrena [4.2K]

Answer: For every one mole of Ca used in this reaction, two mols of H20 are used, one mole of Ca(OH)2 is formed, and one mole of H2 is formed.

Explanation: Once the equation is balanced, you can get the ratio from the coefficients. If you are looking at the ratio of Ca to H2O, the ratio is 1:2; Ca to H2 1:1.

5 0

2 years ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

8 0

3 years ago

The final electron acceptor of the electron transport chain that functions in aerobic oxidative phosphorylation is The final ele

MissTica

Answer:

The final electron acceptor of the electron transport chain is oxygen

Explanation:

Four electrons gotten from cytochrome c are involved in the conversion of a molecule of oxygen (O2) to two molecules of water (H2O). This final electron transfer occurs in complex IV. Complex IV, also known as cytochrome c oxidase, facilitates the the use of four protons from the matrix of the mitochondrion, in the production of water molecules while pumping four protons to the intermembrane space of the mitochondrion.

8 0

3 years ago

WILL AWARD 50 POINTS AND BRAINLIEST THANK YOUUU

adoni [48]

Answer:

chemical symbol= Cl ( chlorine)

electrons= 17

2p electrons = 6

6 0

3 years ago

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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$