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2 years ago

A 50-gram sample has a half-life of 12 days. How much material will remain after 12 days?

2 answers:

8 0

A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life. Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.

Rom4ik [11]2 years ago

7 0

Answer:25 grams

Explanation:

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Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Pepsi [2]

Answer:

$Mass_{chromium}=52.1\ g$

$Mass_{neon}=8.23\times 10^{-7}\ g$

Explanation:

Calculation of the mass of chromium as:-

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1.002\ mol= \frac{Mass}{51.9961\ g/mol}$

$Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g$

Calculation of the mass of neon as:-

Moles = $4.08\times 10^{-8}$ moles

Molar mass of neon = 20.1797 g/mol

Thus,

$1.002\ mol= \frac{Mass}{20.1797\ g/mol}$

$Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g$

6 0

2 years ago

Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the io

Fudgin [204]

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

8 0

2 years ago

Calculate the oxidation number of s in S2O8^2-

mixer [17]

Given problem:

S₂O₈²⁻

Find the oxidation number of S;

Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.

For radicals:

"the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion "

S₂O₈²⁻; oxidation number of O is usually -2

2(S) + 8(-2) = -2

2S - 16 = -2

2S = -2 + 16

2S = 14

S = +7

The oxidation state of S in the radical is +7

3 0

2 years ago

Which pair of compounds has the same empirical formula?

aleksandr82 [10.1K]

i think it's A. cause CH is 1:1 and if you reduce C2H2, the ratio would also be 1:1

5 0

2 years ago

Read 2 more answers

Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be

Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

HCOOH (formic acid) → HCOO⁻ (formate) + H⁺

H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺

Formal Charges:

Formal charges are calculated using following formula,

F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

F.C = [6] - [6 + 2/2]

F.C = [6] - [6 + 1]

F.C = 6 - 7

F.C = -1

For Sodium:

F.C = [1] - [0 + 0/2]

F.C = [1] - [0]

F.C = 1 - 0

F.C = +1

5 0

2 years ago