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stepladder [879]
3 years ago
11

To which period does the Lanthanide series belong? Actinides?

Chemistry
1 answer:
borishaifa [10]3 years ago
7 0
Lanthanide belongs to period 6, and actinides belongs to period 7
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The question is in the picture above ! help!
Crank

Answer:

El pepe

Explanation:

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7 0
3 years ago
How is it that when a salt sample dissolves in water, the delta S for the process is positive?
Bess [88]

Answer:-

Water is highly ordered. In water each oxygen atom is connected to others around it through hydrogen bonding via bridging hydrogen atoms. When a salt like NaCl is dissolved, some of these Hydrogen bonds break.

When a salt like NaCl dissolves in water, the NaCl breaks in to ions Na+ and Cl-.

The water molecules now surround these ions.

The slightly negative oxygen end of water molecule gets near the Na+, while the slightly positive Hydrogen of water molecule gets near the Cl-.

So before salt sample dissolve, the water molecules were highly ordered due to hydrogen bonding. Now after salt dissolve there is a decrease in order and thus an increase in disorder of the water molecules.

Due to increase in disorder, entropy which is a measure of disorder increases. Since entropy increases, delta S for the process is positive.

4 0
3 years ago
All molecules have the same kinetic energy and hence the same speed.<br><br> O True<br><br> O False
topjm [15]
False...their speed differs
4 0
2 years ago
What is pollination?
blagie [28]
Pollination is the act of transferring pollen grains from the male anther of a flower to the female stigma.

Hopefully that helped! Maybe just change up the words a bit <3
5 0
3 years ago
A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber
In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g                 (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532          (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3   = (mass of BaCO3 )/(mass of sample )*100

                                                         = 0.242/(0.846 )*100

                                                         = 28.605%

5 0
3 years ago
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