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Travka [436]
3 years ago
5

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

Chemistry
2 answers:
patriot [66]3 years ago
5 0

Answer:

4

Explanation:

BabaBlast [244]3 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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Which of the following is an example of a covalent bond?
loris [4]

Answer:

C. H2O

yep this one is right

4 0
3 years ago
How many objects are in a mole of objects?
lakkis [162]
Your Question: {How many objects are in a mole?}

Helpful Knowledge: (We Know the amount in an object: 12g or C^12)

{A number of objects that are in a mole of objects?}

Well for the question it is pretty easy to answer because a number of objects in One mole would equal 6.02 × 10²³ 

Which 6.02 × 10²³ is an Avogadro's Number. 

So it depends on how many objects you have.

So for every object you have, One mole would equal 6.02 × 10²³. Or 62,000,000,000,000,0000,000,000. Big Number am I right. So that's why we just use 6.02 × 10²³.

Anywho, your answer would be 6.02 x 10²³ x n. 
N would equal the number of objects you're calculating. 

Final Answer: 6.02 x 10²³ x (n) = (Your Answer)

Hope this helps! Have a great day. If you need anything else, feel free to hope right in my inbox. Or comment below. ↓

6 0
3 years ago
Answer and work for this problem
MArishka [77]
We can write the balanced equation for the synthesis reaction as 
     H2(g) + Cl2(g) → 2HCl(g)

We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
     mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) * 
                           (2.02 g H2 / 1 mol H2)                        
                        = 4.056 g H2

We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
     mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
                             (70.91 g Cl2 / 1 mol Cl2)
                          = 142.4 g Cl2 

Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
6 0
3 years ago
A schoolbus accelerates to 65 mph and enters the freeway. It travels
Inessa [10]

Answer:

Distance cover by school bus = 149.5 miles

Explanation:

Given:

Velocity of school bus = 65 mph

Time taken by school bus = 2.3 hours

Find:

Distance cover by school bus

Computation:

Distance cover = Velocity x Time taken

Distance cover by school bus = Velocity of school bus x Time taken by school bus

Distance cover by school bus = 65 x 2.3

Distance cover by school bus = 149.5 miles

7 0
3 years ago
(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe
Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration C_o = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

time (t) = \dfrac{V}{Q}

time (t) = \dfrac{200}{0.3}

time (t) = 666.66 \ sec

time (t) = \dfrac{666.66 }{3600} hrs

time (t) = 0.185 hrs

\text{Using First Order Reaction:}

\dfrac{dc}{dt}=kc

where;

t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)

0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)

0.942 =  \Big ( \dfrac{200}{C_e}- 1 \Big)

1+ 0.942 =  \Big ( \dfrac{200}{C_e} \Big)

\dfrac{200}{C_e} = 1.942

C_e = \dfrac{200}{1.942}

\mathbf{C_e = 102.98 \ mg/l}

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)

\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)

\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)

V_{PFR} =0.196(0.663)*0.3*3600

V_{PFR} = 140.34 \ m^3

The volume of the PFR is ≅ 140 m³

3 0
3 years ago
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