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NeTakaya
3 years ago
12

Which of the following is the best thermal conductor? metal wood paper air

Chemistry
2 answers:
Ganezh [65]3 years ago
5 0

Answer:

metal

Explanation:

Airida [17]3 years ago
5 0
The answer is to this is metal
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Is NaCl CH3OH LiOH or H2SO4 a bronsted acid
vesna_86 [32]

Answer:

D. H₂SO₄

Explanation:

Bronsted acids are those that donate H+ ions. In this question, H₂SO₄ is a Bronsted acid.

Note: H₂SO₄ is one of seven strong acids that you should try to memorize.

3 0
3 years ago
Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.
N76 [4]

Answer:

1 . Al_2O_3

2. TiO_2

Explanation:

The more stable the ionic compound, the more is it lattice energy.

  • The more the charge on the cation and the anion, the greater is the lattice energy.
  • The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide (Sc_2O_3) is an oxide in which Sc^{3+} behaves as cation and O^{2-} behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . Al_2O_3

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation Al^{3+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

2. TiO_2

This is because the charge on the cation Ti^{4+} is greater than that of Sc^{3+} and also the size of the cation Ti^{4+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

3 0
3 years ago
I’m to lazy can anyone help me please
katen-ka-za [31]

Answer: what do i have to do to help.. ;-; like i need instructions and why are the words coverd off black??

;-;

Explanation:

6 0
3 years ago
If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce
Natali5045456 [20]
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)

=0.0282669621 g of O2 left over
5 0
4 years ago
How much of 45 grams of pu 234 remains after 27 hours if it’s half life is 4.98 hours
nadya68 [22]

Answer:

1.07 g

Explanation:

Half-life of Pu-234 = 4.98 hours

Initially present = 45 g

mass remains after 27 hours = ?

Solution:

Formula

         mass remains = 1/ 2ⁿ (original mass) ……… (1)

Where “n” is the number of half lives

To find "n" for 27 hours

                          n = time passed / half-life . . . . . . . .(2)

put values in equation 2

                          n = 27 hr / 4.98 hr

                          n = 5.4

Mass after 27 hr

Put values in equation 1

          mass remains = 1/ 2ⁿ (original mass)

          mass remains = 1/ 2^5.4 (45 g)

          mass remains = 1/ 42.2 (45 g)

          mass remains =  0.0237 x 45 g

          mass remains =  1.07 g

6 0
4 years ago
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