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Furkat [3]
3 years ago
12

What mass of salt (nacl) should you add to 1.48 l of water in an ice cream maker to make a solution that freezes at -13.4 ∘c ? a

ssume complete dissociation of the nacl and density of 1.00 g/ml for water?
Chemistry
2 answers:
GuDViN [60]3 years ago
7 0

Answer:

311.6g NaCl you should add

Explanation:

When you add a solute (NaCl) to solvent (Water), the freezing point of the solution decreases with regard to pure solvent following the equation:

ΔT = Kf × m × i

Where ΔT is change in temperature(From 0°C to -13.4°C), Kf is freezing point depression constant (1.86°C/m for water), m is molality of solution (Moles solute / 1.48 kg solvent -Because 1.48L≡1.48kg; density 1.00g/mL-) and i is Van't Hoff factor (2 for NaCl because in water, NaCl dissociates as Na⁺ and Cl⁻ ions, 2 ions).

Replacing:

13.4°C = 1.86°C/m × moles NaCl / 1.48kg × 2

5.33 = moles NaCl

As molar mass of NaCl is 58.44g/mol, mass in 5.33moles are:

5.33mol NaCl × (58.44g /mol) = <em>311.6g NaCl you should add</em>

blagie [28]3 years ago
5 0
Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.

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Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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Rudik [331]

Answer:

I think it is C

Explanation:

Hope this helps!! :)

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What type of wave it do the contractions of the snakes muscles make as a snake moves forward
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4 0
3 years ago
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If a patient\'s blood pressure is 145 over 65 mmHg, what is it in atmospheres (atm)
kicyunya [14]

Converting mmHg to atm is solved by division.

Example: Convert 745.0 to atm.

Solution- divide the mmHg value by the 760.0 mmHg / atm.

745 mmHg over 760.0 mmHg/atm

atm value is 0.980263

Now, I am a medical student and we have never had to convert a BP (blood pressure) to atm from mmHg, only ever kPA. SO, I am going to take a guess here and say that when you do the work to solve this, you are going to convert the Systolic (upper #) which is the 145. You should get 0.190789 and then convert the Diastolic (lower #) which is 65. You should get 0.08552632.

So your fraction so to speak should read, 0.190789/0.08552632 or 0.190789 over 0.08552632

(Just to note that is way to low of a BP, although it is irrelevant) Best wishes and good luck. "Remember, never just look for the right answer, look for why it is the right answer!"

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C6H12O6 + 6 O2 --&gt; 6 CO2 + 6 H2O + energy In the reaction listed above, 1 molecule of glucose reacts with 6 molecules of oxyg
kirill115 [55]

Answer:

  • Third choice:<em> energy present in the glucose and oxygen that is not needed for the formation of carbon dioxide and water is released to form energy/ATP.</em>

Explanation:

<u>1) Chemical equation (given):</u>

  • C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O + energy

<u>2) Chemical potential energy:</u>

Each compound stores chemical potential energy.  This energy is stored in the chemical bonds.

Due to every substance has its own unique chemical potential energy, when a chemical reaction takes plase, yielding to the change of some substances, some energy is absorbed (when bonds are formed) and some energy is released (when bonds are broken).

<u>3) Conservation of energy:</u>

Then, if the sum of the bond energies of the final products is less than the sum of the bond energies of the reactants, the<em> law of conservation of energy</em> rules that the difference between the total energies of the products and reactants must be released to the surroundings.

That is what is happening in the given reaction:

  • C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O + energy

The term energy in the product side means that energy is conserved because it is being released due to the the glucose and oxygen (reactant side) have more energy stored in their bonds than the energy needed for the formation of carbon dioxide and water, so that excess of energy is released to form energy/ATP.

<u>Summarizing:</u>

  • The energy on the product side added to the energy of carbon dioxide and water equals the energy of the glucose and oxygen and the final balance is:

  • ∑ Energy of the reactants = ∑energy of the products + released energy, supporting the law of conservation of energy.
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3 years ago
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