A metal bromide, MBr_2, is converted to a molten form at high temperature. Electrolysis of this sample with a current of 6.13A f

The amount to administer to the child is 2,469 mL.
To convert to kilograms (kg), the child's weight in pounds (lb) is multiplied by 0.45359237: m(child) = 72.6 · 0.045359237 = 32.93 kg.
To find m(Medrol), the child's mass in kilograms is multiplied by 1.5 mg/kg.
Thus, m(Medrol) = 32.93 kg · 1.5 mg/kg = 49.39 mg.
The concentration of Medrol is d(Medrol) = 20.0 mg/mL.
To find the volume of Medrol needed, use V(Medrol) = m(Medrol) ÷ d(Medrol).
So, V(Medrol) = 49.39 mg ÷ 20 mg/mL = 2,469 mL.

Answer:

Both CaCl2 and CaBr2 consist of elements (bromine and chlorine) from the same group (group 7).

Explanation:

In the periodic table, elements are arranged into groups based on their valence electron count in the outermost shell. Elements in the same group, which possess a similar number of valence electrons, typically exhibit similar chemical behaviors.

Chlorine and Bromine in CaCl2 and CaBr2 belong to group 7, known as HALOGENS, characterized by having 7 valence electrons in their outer shell.

The similarity in properties between CaCl2 and CaBr2 arises because both contain Chlorine and Bromine, leading to analogous reactions and behaviors when interacting with other compounds.

Response: 1-methoxy-2,4-dinitrobenzene

Rationale:

The nitro groups exhibit a strong electron-withdrawing effect, facilitating nucleophilic substitution reactions where a substituent is replaced by a robust nucleophile like the methoxy group. The reaction's mechanism is illustrated below. The electron-withdrawing nature of the nitro group aids in the formation of the intermediate during the reaction as depicted.

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.

Answer: The process of heating a crucible to eliminate moisture from a hydrate.

Explanation:

The available choices are:

a. Heating a solvent to aid in the dissolution of a solute.

b. Heating a solid in isolation to remove moisture.

c. Bringing water to a boil for use in a water bath.

d. Heating a crucible to eliminate moisture from a hydrate.

Possible actions that can be done on a hot plate include:

a. Heating a solvent to assist a solute in dissolving.

b. Heating a solid in isolation to dry it.

c. Heating water to boiling for a water bath.

However, it's important to note that using a hot plate for heating a crucible to remove water from a hydrate is not advisable. Silica or ceramic materials are not meant to be heated on a hot plate.

Consequently, the correct procedure is heating a crucible to remove water from a hydrate.