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miv72 [106K]
3 years ago
8

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!! DUE TOMORROW 40 POINTS!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
jolli1 [7]3 years ago
8 0

Answer: The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons. They are the electrons involved in chemical bonds with other elements. Every element in the first column (group one) has one electron in its outer shell.

Explanation: :)    

ollegr [7]3 years ago
3 0

Answer: I believe this is the answer: The vertical columns on the periodic table are called groups or families because of their similar chemical behavior. All the members of a family of elements have the same number of valence electrons and similar chemical properties. The horizontal rows on the periodic table are called periods.

Good luck on assignment! :)

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The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74
Alex73 [517]
From  the  equation  above   the  reacting   ratio  of  KClO3   to  O2  is  2:3 therefore  the  number  of  moles  of  oxygen  produced  is  ( 4 x3)/2 =  6 moles  since   four  moles  of  KClO3  was  consumed
mass=relative  formula mass  x  number  of  moles
That  is   32g/mol x 6  moles  =192grams


8 0
3 years ago
Read 2 more answers
if 2.0 g of hydrogen sulfide, H2S(g) reacts with 5.0 g of sodium hydroxide what mass of the excess reactant is present when the
Ronch [10]

Answer: iits 9.g

Explanation:

7 0
3 years ago
A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem
BartSMP [9]

Answer:

Q_{metal} = -6799\,J

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Q_{metal} = - Q_{w}

Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})

Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)

Q_{metal} = -6799\,J

6 0
3 years ago
Read 2 more answers
What is the Bond Type and Molecular polarity of PCl3?
DedPeter [7]
It is covalent bonding. The electrons are shared between the phosphorus and the chlorines.

covalent bonding is when electrons are shared between two elements.

molecular polarity is a little bit complicated, but I will try to explain ;)
PCl3 is an alternation on tetrahedral molecules.
It means that P has one lone pair of electrons. This pair of electrons are only attracted to the P nuclei and thus a greater freedom of motion.
This means that their orbital is bigger and this pushes the 3 Cl atoms closer together.
The angle between each Cl now is 107 and the angle between Cls and P is greater than 107.
Now, due to this shape, and also electronegativity (Cl is more electronegative than P meaning that it tends to hog the electrons they share closer to itself), PCl3 is polar. Electrons that are shared tend to flow closer towards the Cl than the P side.
Therefore, on the Cl side of the molecule it's, more negative. On the P side, it's more positive.
4 0
4 years ago
Read 2 more answers
How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

            no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

                                       = 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0
3 years ago
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