"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O

Answer:

The answer is D. Aluminium Oxide 0.10, Magnesium Oxide 0.50. Firstly, for sodium hydroxide, we can calculate the number of moles using the formula moles = concentration × volume, leading to 0.2 moles from 100 cm³ at 2 moldm-3. Since 2 moles of NaOH yield 1 mole of Al2O3, this means 0.2 moles of NaOH produces 0.1 moles of Al2O3. For hydrochloric acid, moles can similarly be determined, leading to 1.6 moles from 800 cm³ at 2 moldm-3. Considering that 1 mole of Al2O3 reacts with 6 moles of HCl, 0.1 moles of Al2O3 will consume 0.6 moles of HCl. Post-reaction, we have 1 mole of HCl remaining, which will further react with magnesium oxide, thereby yielding 0.5 moles of MgO.

Answer: The correct option is 3.

Explanation: Radioisotopes that emit alpha-particles are termed alpha-emitters. These isotopes undergo alpha-decay.

Those radioisotopes that emit beta-particles are called beta-emitters. They undergo beta-minus decay, in which a neutron converts to a proton and an electron.

Isotopes that emit positrons are known as positron-emitters, undergoing beta-plus decay where a proton becomes a neutron.

From the options given,

Option 1: All three isotopes undergo beta-minus decay.

Option 2: Cs-137 and Tc-99 undergo beta-minus decay.

Fr-220 undergoes alpha-decay.

Option 3: Kr-85 undergoes beta-minus decay.

Ne-19 undergoes positron decay.

Rn-222 undergoes alpha decay.

Option 4: All three isotopes undergo beta-minus decay processes.

Therefore, the correct choice is 3.

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.

The result is 0.14303691.

Carbon-13 (¹³C) is a stable isotope of carbon with a mass number of 13, composed of six protons and seven neutrons.

Isotopes are elements that share the same atomic number but have different mass numbers, meaning they have a varying number of neutrons.

ω(¹³C) = 1.10% ÷ 100%.

ω(¹³C) = 0.0110; this indicates the natural abundance of carbon-13.

m(¹³C) = 13.003355; the atomic mass assigned to carbon-13.

ω(¹³C) · m(¹³C) = 0.0110 · 13.003355.

ω(¹³C) · m(¹³C) = 0.14303691.