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3 years ago

Can Air can be warmed or cooled by the surface below

1 answer:

3 0

I believe it can be warm because usually the surface below us is warm itself, causing the air to be warm as well. The temperature can vary as well

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So2−3(aq)→so2−4(aq) (basic solution) express your answer as a chemical equation. identify all of the phases in your answer. so32

alexira [117]

Answer:

$SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)+2e^{-}$

Explanation:

The given reaction is an example of oxidation reaction. The step-by-step balancing procedure has been shown below:

Chemical equation: $SO_{3}^{2-}(aq.)\rightarrow SO_{4}^{2-}(aq.)$

Balance H and O in basic medium: $SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)$

Balance charge: $SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)+2e^{-}$

Balanced chemical equation:

$SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)+2e^{-}$

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4 years ago

What was the key design change for hfc-134a a/c systems versus CFC 12 a/c systems

BabaBlast [244]

Answer is: key design change for HFC-134a A/C systems versus CFC-12 A/C systems was quick couple service fitting and that design reduce venting and mixing of refrigerants during service.

<span> Level of contamination is also reduced and the emission of refrigerants and greenhouse gases (sulfur dioxide, carbon dioxide) is also reduced.</span>

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3 years ago

When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu

IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

<span>Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.</span>

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3 years ago

2. When paleontologists study fossils, which is true of the fossil record that they find?

juin [17]

Answer:

Explanation:

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3 years ago

Read 2 more answers

How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50

Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

Equilibrium constant (Ka) = 2.3×10¯⁵
pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>

pKa = 4.64
pH = 6.5
Molarity of salt [NaZ] = 0.5 M
Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>

Molarity of HCl = 0.0069 M
Volume = 150 mL = 150 / 1000 = 0.15 L

Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

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2 years ago