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2 years ago

At standard pressure the boiling point of an unsaturated nano3(aq) solution increases when

Chemistry

2 answers:

patriot [66]2 years ago

5 0

Answer:

Explanation:

BabaBlast [244]2 years ago

4 0

Missing question:
(1) the solution is diluted with water.
(2) some of the NaNO3(aq) solution is removed.
(3) the solution is stirred.
(4) more NaNO3(s) is dissolved in the solution.
Answer is: (4) more NaNO3(s) is dissolved in the solution.
Unsaturated solution means that more of a substance can be dissolve (in this case sodium nitrate). If more sodium nitrate is dissolved, molality of solution is increased, eo ipso boiling point of solution.

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Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.

icang [17]

Answer : $3.2 X 10^{15} g/cm^{3}$

Explanation : To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = $1.66054 X 10^{-24} g$

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X $1.66054 X 10^{-24} g/amu$ = $1.673 X 10^{-24} g$

Now, Volume = 1/6πd³ as diameter is given as $1.0 X 10^{-15} m$ converting it to cm will require to multiply with 100

∴ Volume = 1/6π $(1.0 X 10^{-15}mX 100 cm / 1 m)^{3}$

Hence, volume = $5.236 X 10^{-40} cm^{3}$

Therefore, Density = mass / volume

∴ Density = $1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}$

Therefore, Density will be $3.2 X 10^{15} g/cm^{3}$ .

6 0

2 years ago

Read 2 more answers

Calculate the mass of MNO2 needed to produce 25.0 g of Cl2

Vedmedyk [2.9K]

I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O.
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams</span>

7 0

2 years ago

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A 2135 cm3 sample of dry air has a pressure of 98.4 kpa at 127 degrees Celsius. What is the volume of the sample if the Temperat

kykrilka [37]

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2

2135 cm3 divided by 127 degrees celcius = x divided by 206

answer: 3460 cm3

7 0

2 years ago

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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

When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a

AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0

2 years ago