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2 years ago

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Which of the following are oxidation half-reactions?

1 answer:

Olin [163]2 years ago

8 0

The answer is the third choice because is oxidation half reactions, only the oxidation state of the reducing agent changes; in this third choice, Fe2+ is oxidized and becomes Fe3+. Just keep in mind that half reactions usually include the change in electrons:

Fe2+ -> Fe3+ +e-<span />

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The atomic mass of titanium is 47.88 atomic mass units. This atomic mass represents the

kkurt [141]

Answer (3) is the most correct, although (2) is not to be ignored. (3) states the most abundant isotope Ti's average mass, which is certainly true. (2) is the total mass of all protons, neutrons, and electrons in an atom of Ti, which is true but has to be more specific in order to pinpoint exactly the 47.88 amu. (4) is incorrect because it is not of all the naturally occurring isotopes of Ti. (1) is incorrect because they forgot electrons.

3 0

3 years ago

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

kenny6666 [7]

Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

$=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol$

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

5 0

3 years ago

The phase of matter can be changed by:

stiks02 [169]

Answer: A

Explanation: Mass and volume is the amount, amount doesn’t make the phase of a matter change.

3 0

3 years ago

Read 2 more answers

Which of the following factors is the most in-<br> fluential in the formation of soil?

Shtirlitz [24]

B. climate
this is because it influences the speed of chemical reactions in the soil

4 0

3 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago