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3 years ago

Pouring cold water in your hot coffee is conduction convection radiation

Chemistry

2 answers:

Sladkaya [172]3 years ago

6 0

Convection this is because convection is bull movement of molecules between fluids

xxMikexx [17]3 years ago

4 0

Answer:

I would say it is convection

Explanation:

In convection, the heat transfer takes within the fluid.

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Some athletes have as little as 3.0% body fat. If such a person has body mass of 65 kg,how many pounds of body fat does tha

coldgirl [10]

They would have 5%:::::)

8 0

4 years ago

Aldehydes and ketones undergo acid-catalyzed reaction with alcohols to yield hemiacetals, compounds that have one alcohol-like o

REY [17]

Answer:

See explaination

Explanation:

See attachment for the structure of the acetal that you would obtain by reaction of cyclopentanone with 1-propanol.

4 0

3 years ago

The effusion rate of hcl is 43. 2 cm/min in a certain effusion apparatus. What is the rate of effusion of ammonia in the same ap

olga2289 [7]

The rate of effusion of ammonia (NH₃) in the same apparatus is 63.3 cm/min

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the rate of ammonia (NH₃) </h3>

Rate of HCl (R₁) = 43.2 cm/min
Molar mass of HCl (M₁) = 1 + 35.5 = 36.5 g/mol
Molar mass of NH₃ (M₂) = 14 + (3×1) = 17 g/mol

Rate of NH₃ (R₂) =?

R₁/R₂ = √(M₂/M₁)

43.2 / R₂ = √(17 / 36.5)

Cross multiply

43.2 = R₂ × √(17 / 36.5)

Divide both side by √(17 / 36.5)

R₂ = 43.2 / √(17 / 36.5)

R₂ = 63.3 cm/min

Thus, the rate of effusion of ammonia is 63.3 cm/min

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

5 0

2 years ago

Ayo that was cool, ya'll amazing. <br><br> Another one. Next one gonna be 10 points.

Arisa [49]

Person above is correct

7 0

3 years ago

A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express

Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

$wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63$

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

$V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w$

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

$\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3$

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

$Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3$

We can now calculate the molarity as

$Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}$

8 0

3 years ago