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katovenus [111]
3 years ago
15

Technetium-104 has a half life of 18.0 minutes. how much of a 165.0 g sample remains after 90.0 minutes?

Chemistry
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

UkoKoshka [18]3 years ago
4 0

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

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An ionic bond occurs between what particles?
atroni [7]

Answer:

A positive and negative ion.

Explanation:

Ionic bond:

It is the bond which is formed by the transfer of electron from one atom to the atom of another element.  

Both bonded atoms have very large electronegativity difference.

The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion.  Sodium have one valance electron while chlorine have 7 valance electrons. In order to complete the octet chlorine require one electron while sodium need to lose its one electrons. That's why when both atom combine sodium lose its electron and becomes positive ion i.e cation while chlorine accept its electron and becomes negative ion called anion and bond between them is ionic bond.

4 0
3 years ago
Read 2 more answers
How many grams of MgCl2 (molar mass = 95.20 g/mol) will be formed from 25.6 mL of a 0.100 M HCL solution reacting with excess ma
lorasvet [3.4K]
The balanced equation for the above reaction is as follows;
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of HCl to MgCl₂ is 2:1
we have been told that Mg is in excess therefore HCl is the limiting reactant 
number of HCl moles reacted - 0.100 mol/L x 0.0256 L = 0.00256 mol
according to molar ratio, number of MgCl₂ moles formed - 0.00256/2 
Therefore number of MgCl₂ moles formed - 0.00128 mol
mass of MgCl formed - 0.00128 mol x 95.20 g/mol = 0.122 g
3 0
3 years ago
Read 2 more answers
What mass of H₂ is needed to react with 8.75 g of O₂ according to the following equation: O2(g) + H2(g) → H₂O(g)?
FromTheMoon [43]

Explanation:

For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

The given balanced equation has been:

\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O

From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.

The mass of oxygen has been: 8.75 g,

Moles = \rm \dfrac{weight}{molecular\;weight}molecularweightweight

Moles of oxygen = \rm \dfrac{8.75}{32}328.75

Moles of oxygen = 0.27 mol

Since,

1 mole Oxygen = 2 mole hydrogen

0.21 mol oxygen = 0.54 mol hydrogen

Mass of hydrogen = moles \times× molecular weight

Mass of hydrogen = 0.54 \times× 2

Mass of hydrogen = 1.08 grams.

Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

6 0
1 year ago
What is the pH of 0.0001 M NaOH
Gnom [1K]
The pH of NaOH is 10
6 0
3 years ago
Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h
balu736 [363]

CxHy     +  O2    -->    x CO2     +    y/2  H2O

 

Find the moles of CO2 :     18.9g  /  44 g/mol   =    .430 mol CO2   = .430 mol of C in compound

Find the moles of H2O:      5.79g / 18 g/mol     =     .322 mol H2O   = .166 mol of H in compound

 

Find the mass of C and H in the compound:

                             .430mol  x 12  =  5.16 g C

                              .166mol  x 1g   = .166g H   

 

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

8 0
3 years ago
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