1) The metal which reduces the other compound is the one higher in the reactivity. So in this case it is 
.
2) The substance which brings about reduction while itself getting oxidised (that is losing electrons) is called a reducing agent. Here, $\mathrm{Zn}$ is the reducing agent and reduces Cobalt Oxide to Cobalt while itself getting oxidised to Zinc oxide.
P1 * V1 ÷ T1 = P2 * V2 ÷ T2
45 * 1.20 ÷ 314 = 96 * V2 ÷ 420
30,144 * V2 = 22,680
V2 = 22,680 ÷ 30,144
The new volume is approximately 0.75 liter.
I hope I helped
Expanding in powers 4FCsixteen.
<h3>What is hexadecimal system?</h3>
Hexadecimal is an easy route for addressing twofold. It is vital to take note of that PCs don't utilize hexadecimal - it is utilized by people to abbreviate parallel to an all the more effectively reasonable structure. Hexadecimal Number System is generally utilized in Computer programming and Microprocessors. It is likewise useful to portray colors on pages. Every one of the three essential tones is addressed by two hexadecimal digits to make 255 potential qualities, hence bringing about in excess of 16 million potential tones. The primary benefit of utilizing Hexadecimal numbers is that it utilizes less memory to store more numbers, for instance it stores 256 numbers in two digits while decimal number stores 100 numbers in two digits. This number framework is likewise used to address Computer memory addresses.
Learn more about hexadecimal system, refer:
brainly.com/question/21751836
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Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
Answer:
2.29 moles of Cr₂O₃ are produced
Explanation:
This is the reaction:
4 Cr + 3O₂ → 2Cr₂O₃
Ratio for this equation is 4:2, so 4 moles of chromium can produce the half of moles of chromium(III) oxide
4.58 mol of Cr may produce (4.58 .2)/4 = 2.29 moles of Cr₂O₃
