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Feliz [49]
3 years ago
10

What is 2-2? When the cow is green

Chemistry
2 answers:
bija089 [108]3 years ago
6 0
0 with a green cow lol
skad [1K]3 years ago
5 0
0 and the cow remains green (:
You might be interested in
An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
3 years ago
Balance the equations.<br> Zn+ HCl →<br> ZnCl2 +
motikmotik

Answer:

Zn + 2HCl → ZnCl2 + H2

Explanation:

Zn + HCl → ZnCl2 +

The complete equation is given below:

Zn+ HCl → ZnCl2 + H2

Now we can balance the equation by doing the following:

There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:

Zn + 2HCl → ZnCl2 + H2

7 0
3 years ago
Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a
lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

  • Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT
  • Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g
  • Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT
  • Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g
  • Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

7 0
3 years ago
Nitrogen has the atomic number 7. an isotope of nitrogen containing seven neutrons would be
BigorU [14]
An isotope of nitrogen containing 7 neutrons would be nitrogen-7
8 0
3 years ago
An aqueous solution is found to be 3.64% potassium nitrate by mass, how many grams of solution should I evaporate the water from
Sidana [21]

Answer:

513.74 g of solution

Explanation:

% Mass grams are defined as the <em>grams that are dissolved in salt</em> (in this case, it would be <em>potassium nitrate</em>) <em>dissolved every 100 g of the solution</em>. Having this information, you can calculate the amount of solution that has dissolved 18.7 g of potassium nitrate, which is what we want to obtain.

The relationship is:

3.64 g of potassium nitrate _____ 100 g solution

18.7 g of potassium nitrate _____ X = 513.74 g of solution

Calculation: 18.7g x 100g / 3.64g = 513.74 g of solution

So, <em>I need 513.74 g of solution to get 18.7g of potassium nitrate by evaporating it</em>.

8 0
2 years ago
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