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yawa3891 [41]
2 years ago
5

Please help need to turn in

Chemistry
2 answers:
ohaa [14]2 years ago
8 0

Answer:

2. d

3. i

4. e

5. j

6. g

7. a

8. f

9. b

Explanation:

do a few searches when you have time to deepen understanding!

nignag [31]2 years ago
7 0
5. j
6. g
4. e
9. b
2. d
7. a
10. h
1. c
3. f
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How many grams (mass) are in 1.7 moles of Nickel II oxide NiO?
WARRIOR [948]

Answer:

126.99g

Explanation:

NiO has a molar mass of 74.7 g/mol. This means that one mole of nickel (II) oxide has a mass of 74.7 grams. We can use stoichiometry/dimensional analysis to figure out how many grams are in 1.7 moles.

1.7mol NiO (\frac{74.7grams}{1 mole} ) = 126.99g

3 0
3 years ago
Read 2 more answers
Which element requires the least i.e. potassium or gallium why
-Dominant- [34]

Answer:

Potassium

General Formulas and Concepts:

<u>Chem</u>

  • Reading a Periodic Table
  • Periodic Trends
  • Ionization Energy - energy required to remove an electron from a given element
  • Coulomb's Law
  • Shielding Effect
  • Z-effective and Forces of Attraction

Explanation:

The Periodic Trend for 1st Ionization Energy is increasing up and to the right. That means He would have the highest I.E and therefore take the most amount of energy to remove an electron.

Potassium and Gallium are both in Period 4. Potassium is element 19 and Gallium is element 31.

Potassium's electron configuration is [Ne] 4s¹ and Gallium's electron configurations is [Ne] 4s²3d¹⁰4p¹. Since both are in Period 4, they have the same number of core e⁻. Therefore, the shielding effect is the same.

However, since Gallium is element 31, it has 31 protons compared to Potassium, which is element 19 and has 19 protons. Gallium would have a greater Zeff than Potassium as it has more protons. Therefore, the FOA between the electrons and nucleus of Ga is much stronger than that of K. Thus, Ga requires <em>more</em> energy to overcome those FOA to remove the 4p¹ e⁻. Since K has less protons, it will have a smaller Zeff and thus less FOA between the e⁻ and nucleus, requiring <em>less</em> energy to remove the 4s¹ e⁻.

7 0
3 years ago
What type of product forms in the intramolecular reaction between the aldehyde portion of the glucose molecule below and its C-5
djyliett [7]

Question:

What type of product forms in the intramolecular reaction between the aldehyde portion of the glucose molecule below and its C-5 hydroxyl group?  

a. disaccharide

b. carboxylic acid

c. hemiacetal

d. ester

e. stereoisomer

Answer:

hemiacetal  forms in the intramolecular reaction between the aldehyde portion of the glucose molecule and its C-5 hydroxyl group

Explanation:

It is an alcohol also an ether that has been attached to the carbon molecule. Here the hydrogen has occupied the fourth bonding position. This hemiacetal has been derived from the aldehyde. Hence, hemiketal being an alcohol as well as ether has been attached to the same carbon and also to the two other carbon.

7 0
3 years ago
How is chemistry used in chemical engineering?
Dmitriy789 [7]

Answer: Chemical engineers use chemistry and engineering to turn raw materials into usable products, such as medicine, petrochemicals, and plastics on a large-scale, industrial setting. They are also involved in waste management and research. They may be involved in designing and constructing plants as a project engineer.

Explanation:

4 0
3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135

8 0
3 years ago
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