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3 years ago

How do i find the mass of oxygen in 250 grams of sulfuric acid?

1 answer:

6 0

250g H2SO4 x 1moleH2SO4/ 98g mole H2SO4 x 4moles Oxygen/1moleH2SO4 x 1moleOxgen/16g Oxygen
= 163.27g Oxygen

my figures aren't exact....so it's 160g

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1. Which of the following particles are free to drift in metals?

kotegsom [21]

Answer:

electrons

Explanation:

metals do what is called metallic bond

8 0

2 years ago

A student made the Lewis dot diagram of a compound as shown. Mg is written with two dots shown on its top. Cl is written on the

blsea [12.9K]

Answer:

The dots were not properly located and arrows are not used in Lewis structures

Explanation:

If we intend to write a Lewis structure for a compound, that lewis structure must consist of only dots. These dots actually show the valence electrons on the outermost shell of the molecule.

We do not involve arrows when writing dot electron structures for compounds. The valence electrons of magnesium ought not to be written together because they are not a lone pair, rather they are two unpaired electrons. The use of an arrow suggests a coordinate covalent bond which is not the case here.

The correct lewis structure for MgCl2 is shown in the image attached to this answer.

7 0

3 years ago

How many molecules are there in 3 moles of aluminum chloride ?

Svetradugi [14.3K]

there are 4 atoms in aluminum chloride

6 0

2 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

2 years ago

Complete the mechanism for the base-catalyzed opening of the epoxide below by adding any missing atoms, bonds, charges, nonbondi

worty [1.4K]

Complete Question:

check the first image for complete part of the question

Answer and Explanation:

Epoxide is a three membered ring made up of two carbon atoms and one oxygen atom. Epoxides are cyclic ethers. Due to its ring size, it is highly strained and very reactive. Epoxide ring opening takes place with respect to addition of acid and base.

Ring opening of epoxide with acid:

In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)

Ring opening of epoxide with base:

The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.

(check file 2 attached)

3 0

2 years ago