What will happen when a piece of magnesium metal is dropped into a beaker containing a l M solution of copper(1) chloride?

A flexible plastic container contains 0.860g of helium has in a volume of 19.2L if 0.205g of helium is removed at contact pressu

mash [69]

Given:

Initial volume of He, V1 = 19.2 L

Initial mass of He, m1 = 0.0860 g

Mass of He removed = 0.205 g

To determine:

The new volume of He i.e V2

Explanation:

Based on Avogadro's law:

Volume of a gas is directly proportional to the # moles of the gas

Volume (V) α moles (n) -----(1)

Atomic mass of He = 4 g/mol

Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles

Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles

Based on eq(1) we have:

V1/V2 = n1/n2

V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L

Ans: New volume is 14.6 L

6 0

2 years ago

Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.

icang [17]

Answer : $3.2 X 10^{15} g/cm^{3}$

Explanation : To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = $1.66054 X 10^{-24} g$

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X $1.66054 X 10^{-24} g/amu$ = $1.673 X 10^{-24} g$

Now, Volume = 1/6πd³ as diameter is given as $1.0 X 10^{-15} m$ converting it to cm will require to multiply with 100

∴ Volume = 1/6π $(1.0 X 10^{-15}mX 100 cm / 1 m)^{3}$

Hence, volume = $5.236 X 10^{-40} cm^{3}$

Therefore, Density = mass / volume

∴ Density = $1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}$

Therefore, Density will be $3.2 X 10^{15} g/cm^{3}$ .

6 0

2 years ago

Read 2 more answers

Suppose that a metal oxide of formula m2o3 were soluble in water. what would be the major product or products of dissolving the

V125BC [204]

Meta oxides are compounds that are formed by reaction of metals with oxygen. If these compounds are placed in water, the ionic components of this substance will dissociate.

The dissociation of metal oxides in water will likely form,

2M³⁺ + 3O²⁻

6 0

2 years ago

Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right

svetlana [45]

Answer:

B = CHCl2 + Cl2 --> CHCl3 + Cl

Explanation:

Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.

Free radical chlorination is divided into 3 steps which are:

The initiation step

The propagation step

The termination step

So in reference to the question, propagation step involves two steps.

The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.

The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.

You would find in the attachment the 2 step mechanism.

3 0

2 years ago

Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G

AnnyKZ [126]

1) Chemical equation

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) molar ratios

1 mol Cu: 2 moles AgNO3 : 1 mol Cu (NO3)2 : 2 mol Ag

3) Convert 12. 83 * 10^23 atoms of Cu in moles

12.83 * 10 ^ 23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Use the proportions

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) Use the atomic mass of silver to convert 4.262 mol in grams

mass = number of moles * atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g

5 0

2 years ago