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tester [92]
3 years ago
15

Metallic properties tend to increase in which direction on the periodic table

Chemistry
1 answer:
tresset_1 [31]3 years ago
3 0
Metallic properties head to the left.
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Predict the product(s) of the following neutralization reaction. Type the balanced the equation.
REY [17]

Answer:

The products from the neutralization reaction are barium nitrate and water.  

The balanced chemical equation is

<u>2</u> HNO_{3} +  <u>1</u> Ba(OH)_{2}  -->  <u>1</u> Ba(NO_3)_{2} +  <u>2</u> H_{2}O.  

6 0
3 years ago
Arrange the following in order of increasing bond strength of the carbon oxygen bond: Group of answer choices Carbon monoxide &l
Irina-Kira [14]

Answer:

carbonate ion < carbon dioxide < carbon monoxide

Explanation:

Bond strength depends on the bond order of a bond. The higher the bond order, the greater the bond strength since shorter bonds are stronger than longer bonds.

The carbonate ion has a bond order of 1.33, carbon dioxide has a bond order of 2 while carbon monoxide has a bond order of 3.

Since the compound with the highest C-O  bond order has the strongest C-O bond, then carbon monoxide possesses the strongest C-O bond.

3 0
3 years ago
Choose the options below that are true of a solution of a solid in a liquid. (select all that apply) Select all that apply: Most
Alekssandra [29.7K]

Answer:

Most solids in solution exhibit a general trend of increasing solubility with increasing temperature.

A seed crystal may be added to a supersaturated solution to precipitate excess solute.

Explanation:

For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions(Lumen Learning).

When a seed crystal is added to a supersaturated solution, excess solute begin to precipitate because  the seed crystal now furnishes the required nucleation site where the excess dissolved crystals now begin to grow.

6 0
3 years ago
At high temperatures, carbon reacts with O2 to produce CO as follows: C(s) O2(g) 2CO(g). When 0.350 mol of O2 and excess carbon
777dan777 [17]

Answer:

Value of K_{c} is 0.090.

Explanation:

Initial molarity of O_{2} = \frac{0.350}{5.00}M = 0.0700 M

Construct an ICE table corresponding to the combustion reaction of carbon to determine K_{c}

                       C(s)+O_{2}(g)\rightarrow 2CO(g)

              I (M):    -      0.0700        0

             C (M):  -          -x             +2x

             E (M):   -    0.0700-x       2x

So, K_{c}=\frac{[CO]^{2}}{[O_{2}]}  , where [CO] and [O_{2}] represents equilibrium concentration of CO and O_{2} respectively.

Here, [CO]=2x=0.060

       ⇒x = 0.030

So, [O_{2}] = 0.0700-x = (0.0700-0.030) = 0.040

Hence,  K_{c}=\frac{(0.060)^{2}}{0.040}=0.090

4 0
4 years ago
Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B
Ivenika [448]

Answer:

-2546 kJ

Explanation:

It is possible to obtain the enthalpy of a reaction from the sum of different intermediate reactions.

For the reaction:

4B(s) + 3O₂(g) → 2B₂O₃(s)

The intermediate reactions are:

A- B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g), ΔH°A= +2035 kJ

B- 2B(s) + 3H₂(g) → B₂H₆(g), ΔH°B= +36 kJ

C- H₂(g) + 1/2O₂(g) → H₂O(l), ΔH°C= -285 kJ

D- H₂O(l) → H₂O(g), ΔH°D= +44 kJ

2B = 4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH°2B= +78 kJ

-2A = 6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH°-2A= -4070 kJ

-6C = 6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH°-6C= +1710 kJ

-6D = 6H₂O(g) → 6H₂O(l) ΔH°-6D = -264 kJ

The sum of 2B - 2A - 6C - 6D produce:

4B(s) + 3O₂(g) → 2B₂O₃(s)

And the enthalpy is: ΔH°2B + ΔH°-2A + ΔH°-6C + ΔH°-6D = <em>-2546 kJ</em>

I hope it helps!

3 0
3 years ago
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