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boyakko [2]
3 years ago
7

A warm front moves into a region. What kind of weather most likely results?

Chemistry
2 answers:
Elanso [62]3 years ago
7 0

Answer:

B

Explanation:

it depends if the weather was cold but ummm they warm and cold air would mix and turn story and watch out tornadoes could also happen

Mila [183]3 years ago
6 0

Answer:

Stormy

Explanation:

Apex

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Which would increase the amount of dissolved oxygen in a pond
AlexFokin [52]

Answer:

Dissolved oxygen levels are increased by supplementing wind and wave action, adding plants to water and exposing water to purified oxygen. Also removing fish could increase the amount of dissolved oxygen

Explanation:

6 0
3 years ago
For the following, answer Acid, Base, Both, or Neither
Fynjy0 [20]

Answer:

1. Both

2. Acid

3. Acid

4. Base

5. Acid

6. Neither

7. Neither

8. Base

9. Acid

Explanation:

In Chemistry, a chemical compound can be either a base or an acid.

An acid reacts with metals to produce bubbles of hydrogen gas and it also reacts with carbonates while a base feels slimy to the touch. They both can change the color of a litmus paper.

Base has a sour taste while acids have a bitter taste (please do not use this characteristic to test for an acid in the laboratory)

6 0
3 years ago
What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Ass
Nataly [62]

Answer:

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

Explanation:

Step 1: Data given

the contribution of protons from H2SO4 is near 100 %.

Step 2: The unbalanced equation

.Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + H2O(l)

Step 3: Balancing the equation

On the left side we have 4x H (2x in Ba(OH)2 and 2x in H2SO4). On the right side, we have 2x H (in H2O).

To balance the amount of H on both sides, we have to muliply H2O (on the right side) by 2.

Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

Step 4: The net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

After canceling those spectator ions in both side, look like this:

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

4 0
3 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
4 years ago
Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
muminat

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

<em>-802.3kJ</em>

7 0
3 years ago
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