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3 years ago

How calculate how many milliliters of glycerin (specific gravity=1.26) will have a mass of 0.75 lbs?

1 answer:

5 0

Specific gravity is the ratio between the density of the material to the density of water at 4 degrees celcius (density of water at this temperature is 1000 kg/m^3)

1.26 = density of glycerin / 1000
density of glycerin = 1260 kg/m^3

1 Kg is equivalent to 2.2 lbs, therefore:
0.75 lbs = 0.34 kg

density = mass/volume
1260 = 0.34/volume
volume = 0.34/1260 = 2.857 x 10^-4 m^3

now, 1 m^3 is equivalent to 1000 liters
2.857 x 10^-4 m^3 = 0.2857 liters

one liter is equivalent to 1000 ml
therefore,
0.2857 l = 285.7 ml

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3 years ago

In which of the following reactions does a decrease in the volume of the reaction vessel at constant

Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

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3 years ago

Which equation can be used to convert from Celsius to Kelvin? O Celsius + 459.67 O (Celsius)x1.8+32 O (Celsius-32)x5/9 Celsius+2

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Answer:

Celsius +273.15

Explanation:

The two scales have the same size degree, but Kelvin is an absolute scale based on absolute zero while 0° in Celsius is based on the melting point of water. So, in order to convert from Celsius degrees to Kelvin we only need to add 273.15 to the given temperature:

K= °C + 273.15

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3 years ago

Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends

IgorLugansk [536]

24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its kinetic energy.

$E_\text{k} = 1/2 \; m \cdot v^{2}$

Where

$E_\text{k}$ is the kinetic energy of the molecule,
$m$ its mass, and
$v^{2}$ the square of its speed.

Besides, the kinetic theory of gases suggests that for an ideal gas,

$E_\text{k} \propto T$

where $\text{T}$ its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the average kinetic energy of molecules shall be the same for any ideal gas at the same temperature. So is the case for HCl and NH₃

$E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)$

$m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

Where

$m(\text{HCl})$ , $v(\text{HCl})$ , and $E_\text{k}(\text{NH_3})$ the mass, speed, and kinetic energy of an HCl molecule;
$m(\text{NH}_3)$ , $v(\text{NH}_3)$ , and $E_\text{k}(\text{NH}_3)$ the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their molar mass. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, $m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3)$ . Therefore:

$36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

$36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)$

$\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)$

The average speed NH₃ molecules would be $\sqrt{36.45/17.03} \approx 1.463$ if the average speed of HCl molecules $v(\text{HCl})$ is 1.

$\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}$

$\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}$

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3 years ago

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Answer:

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