There are various kind of elements that are present in periodic table. Elements that are metals are kept on left side of periodic table and elements which are non metals are kept on right side of periodic table.
Periodic table is a table in which we find elements with properties like metals, non metals and metalloids element arranges in increasing atomic number.
The first 20 elements of periodic table are Hydrogen, Helium, Lithium, Beryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Sodium, Magnesium, Aluminum, Silicon, Phosphorus, Sulfur, Chlorine, Argon, Potassium and Calcium. Out of these Lithium, Beryllium, Sodium, Magnesium. Potassium and Calcium are metals so they are kept on the left side of the periodic table while others are non metals so they are kept on right side of periodic table.
Learn more about periodic table, here:
#SPJ2
Answer:
(C3H8) produces 660 g of CO2 and 360 g of H2O
Explanation:
The balanced chemical equation for the combustion of propane (C3H8) is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
This equation tells us that for every molecule of propane (C3H8) that reacts with 5 molecules of oxygen (O2), 3 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O) are produced.
So, if we have 220. g of propane (C3H8), we can find the amount of CO2 and H2O produced by using the mole ratio from the balanced equation:
1 mole C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O
We can find the number of moles of C3H8 by dividing the mass by the molar mass of C3H8 (44 g/mol):
220 g / 44 g/mol = 5 moles C3H8
So, the number of moles of CO2 and H2O produced can be found by multiplying the number of moles of C3H8 by the mole ratio:
3 moles CO2 = 3 moles CO2/1 mole C3H8 * 5 moles C3H8 = 15 moles CO2
4 moles H2O = 4 moles H2O/1 mole C3H8 * 5 moles C3H8 = 20 moles H2O
Finally, we can convert the number of moles of CO2 and H2O to grams by multiplying by their molar masses (44 g/mol for CO2 and 18 g/mol for H2O):
15 moles CO2 * 44 g/mol = 660 g CO2
20 moles H2O * 18 g/mol = 360 g H2O
So, the combustion of 220 g of propane (C3H8) produces 660 g of CO2 and 360 g of H2O.
___2.Variables
___3.Conclusion
___4.Scientific Method
___5.Procedure
a.The steps you take to complete the experiment
b.Factors that changes in an experiment
c.A possible solution to a problem
d.The result of the experiment
e.The process scientist follow to complete an investigation
Answer:
1. d
2. b
3. d
4. e
5. a
explanation:
there's nothing else to explain
B. Quantitive observation
C. Evaluation
D. Qualitative observation
Answer:
B (Quantitative)
Explanation:
a clever way to remember the difference between quantitative and qualitative is that there is a nin quantitative which is the first letter to the word number.So quantitative is an oberservation with numbers
B. 12.43 L
C. 4.77 L
D. 3.35 L
Answer:
Option A. 2.82 L
Explanation:
Step 1:
Data obtained from the question.
pH = 13.55
Step 2:
Determination of the pOH of the solution.. This is illustrated below:
pH + pOH = 14
pH = 13.55
13.55 + pOH = 14
Collect like terms
pOH = 14 - 13.55
pOH = 0.45
Step 3:
Determination of the concentration of the OH ion.
This is illustrated below:
pOH = - Log [OH-]
pOH = 0.45
0.45 = - Log [OH-]
- 0.45 = Log [OH-]
[OH-] = antilog (- 0.45)
[OH-] = 0.355 M
Step 4:
Determination of the molarity of KOH. This is illustrated below:
First, we'll write the dissociation equation of KOH as follow:
KOH —> K+(aq) + OH-(aq)
From the balanced equation above,
1 mole of KOH produced 1 mole of OH-.
Therefore, 0.355 M of KOH will definitely produce 0.355 M of OH-.
The molarity of KOH is 0.355 M
Step 5:
Determination of the volume of the solution needed to dissolve 1 mole of KOH. This is illustrated below:
Mole of KOH = 1 mole
Molarity of KOH = 0.355 M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 1/0.355 M
Volume = 2.82L
0.0340 g O2
Step 1. Write the balanced chemical equation
4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3
Step 2. Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]
= 4.250 mmol Fe^(2+)
Step 3. Calculate the moles of O2
Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]
= 1.062 mmol O2
Step 4. Calculate the mass of O2
Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2
= 0.0340 g O2
0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.
Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol
According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:
Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2
Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):
Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2
Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
Learn more about precipitate the iron here:
#SPJ6