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2 years ago

In a reaction, 25 grams of reactant

Chemistry

1 answer:

Vsevolod [243]2 years ago

8 0

Answer:

D. 15g

Explanation:

The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.

In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.

Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.

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30. how many grams of boric acid, b(oh)3 (fm 61.83), should be used to make 2.00 l of 0.050 0 m solution? what kind of fl ask is

grigory [225]

Answer:

Mass = 6.183 g

Solution:

Step 1: Calculate number of moles of Boric acid using following formula,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 0.05 mol.L⁻¹ × 2.0 L

Moles = 0.1 mol

Step 2: Calculate Mass of Boric Acid using following formula,

Moles = Mass ÷ M.mass

Solving for Mass,

Mass = Moles × M.mass

Putting values,

Mass = 0.1 mol × 61.83 g.mol⁻¹

Mass = 6.183 g

Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.

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2 years ago

Gina wants to use models to better understand how the types of bonds in a molecule relate to the presence of geometric isomers.

vazorg [7]

The answer to this question is D! The ball and stick model! Hope this helps :)

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1 year ago

Read 2 more answers

What is the volume, in liters, occupied by 1.73 moles of N2 gas at 0.992 atm pressure and a temperature of 75º C? (R value- 0.08

taurus [48]

Volume of the nitrogen gas = 49.8 L

Explanation:

It is given that the pressure, number of moles and temperature of nitrogen gas, and gas constant value being constant and it is taken as 0.08206 L atm mol⁻¹K⁻¹.

Temperature = T = 75°C = 75 + 273 = 348 K

Pressure = P = 0.992 atm

Number of moles = n = 1.73 moles

We have to use the ideal gas equation, PV = nRT, and rearranging the equation to get Volume in litres.

V = $$\frac{nRT}{P}$

= $$\frac{1.73\times 0.08206\times348}{0.992}$

= 49.8 L

So the volume of Nitrogen gas = 49.8 L

7 0

1 year ago

The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C

coldgirl [10]

Answer:

The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁.

Explanation:

Particles A and C are shown as if they are on the same vertical line, which means with the same kinetic energy. Both particle A and C are to the lett of particle B, which means that the formers have a lower kinetic energy than the latter.

Since the likelyhood of a particle to participate in the reaction increases with the kinetic energy, particle B is more likely to participate in the reaction than particles A and C. Hence, the first choice is incorrect.

The graph, although not perfectly symmetrical, does show a bell shape, hence there are many particles will low kinetic energy and many particles with high kinetic energy. You cannot assert that most of the particles of the two gases have high high speeds. Hence, second statement is incorrect, too.

At high values of kinetic energy (toward the right of the curve), the line labeled T₁ is higher than the line labeled T₂, meaning that at T₁ more particles have an elevated kinetic energy than the number of particles that have an elevated kinetic energy at T₂.

On the other hand, at low values of kinetic energy (toward the left of the curve) the line T₂ is higher than the line T₁, meaning that at T₂ more particles have a low kinetic energy than the number of particles that have low kinetic energy at T₁.

Hence, the last two paragraphs are telling that the average kinetic energy of gas particles at T₂ is is lower than the average kinetic energy of gas particles at T₁.

Since the average speed is proportional the the square root of the temperature, the same trend for the average kinetic energy is true for the average speed, and you conclude that the last statement is true: "The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁".

Since more particles at T₁ have high kinetic energy than the number of particles at T₂ that have a high kinetic energy, more particles of gas at T₁ are likely to participate in the reaction than the gas at T₂, and the third statement is incorrect.

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1 year ago

Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to

Alexandra [31]

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode $(m_{ch})$ is given as :

$\frac{m_{ch}}{M} ZF = It$

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; $(m_{ch})$ = $(Ad) \rho$ ; replacing that into above equation; we have:

$\frac{(Ad) \rho}{M} ZF = It$ ---- equation (1)

where;

A = area

d = thickness

$\rho$ = density

From the above equation (1); The time required for coating can be calculated as;

$[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t$

$t = \frac{2100}{20}$

= 105 s

8 0

2 years ago