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Gnesinka [82]
2 years ago
5

When some of the sugar added to iced tea remains undissolved at the bottom of the glass, the solution is ________.?

Chemistry
1 answer:
Degger [83]2 years ago
7 0
When some of the sugar added to iced tea remains undissolved at the bottom of the glass, it is the <span>saturated solution </span>at this temperature.
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Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 ∘C? (HINT: the act
svp [43]

Explanation:

Below is an attachment containing the solution.

3 0
2 years ago
A 45 mL sample of nitrogen gas is cooled from 135ºC to 15C in a container that can contract or expand at constant pressure. Wha
Vanyuwa [196]

Answer:

V₂ =31.8 mL

Explanation:

Given data:

Initial  volume of gas = 45 mL

Initial temperature = 135°C (135+273 =408 K)

Final temperature = 15°C (15+273 =288 K)

Final volume of gas = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 45 mL × 288 K / 408 k

V₂ = 12960 mL.K / 408 K

V₂ =31.8 mL

8 0
2 years ago
Balance the following oxidation-reduction reaction: Fe(s)+Na+(aq)→Fe2+(aq)+Na(s) Express the coefficients as integers separated
aliina [53]

Answer:

The answer to your question is: 1, 2, 1, 2

Explanation:

                       1 Fe(s)  + 2 Na⁺(aq)  → 1 Fe²⁺(aq)  + 2 Na(s)

                             Fe⁰   -   2e⁻       ⇒           Fe⁺²        Oxidases

                             Na⁺   +  1 e⁻       ⇒           Na⁰         Reduces

                      1 x ( 1 Fe⁰      ⇒         1 Fe⁺²)      Interchange number of

                      2 x ( 2Na⁺       ⇒       2 Na⁰ )      electrons

6 0
2 years ago
The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close
Natasha_Volkova [10]

Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log(A^-/HA) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

6 0
2 years ago
Where in a data table should units of measurement be shown?
Vikentia [17]
The unites of measurement in a data table should be shown in the headings of some columns.
6 0
2 years ago
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