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3 years ago

3. Nicotine, the addictive drug in cigarettes, contains 74.0% carbon, 8.6% hydrogen,

Chemistry

1 answer:

Vinvika [58]3 years ago

8 0

Answer:

the mass of each element can be recovered from is 162.23 grams

Explanation:

the mass really depends on the volume. If there is 74.0% carbon, 8.6% hydrogen and 17.3% nitrogen then you first cross out the 0 and 3 which leaves you with 17% and 74%. Then all you have to do is multiply 17% and 74% and then divided it by 8.6%.

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How many 3-liter balloons could the 12-L helium tank pressurized to 160 atm fill? Keep in mind that an "exhausted" helium tank i

vaieri [72.5K]

Answer:

The 12L helium tank pressurized to 160 atm will fill 636 3-liter balloons

Explanation:

It is possible to answer this question using Boyle's law:

$P_1V_1=P_2V_2$

Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:

160atm×12L = 1atm×V₂

V₂ = 1920L

As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:

1920L - 12L = 1908L

1908L will fill:

1908L× $\frac{1balloon}{3L}$ = 636 balloons

I hope it helps!

7 0

3 years ago

Can anybody define hypothesis in their own words? (its for school)

Trava [24]

Answer:

An educated guess based on what you already know.

Explanation:

6 0

3 years ago

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What is a cone of depression

Sav [38]

A cone of depression occurs in an aquifer when groundwater is pumped from a well. In an unconfined aquifer, this is an actual depression of the water levels. In confined aquifers, the cone of depression is a reduction in the pressure head surrounding the pumped well.

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3 years ago

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What would the volume of gas be at 150 c if it had a volume of 693 ml at 45 c

marissa [1.9K]

Answer:

$\boxed{\text{922 mL}}$

Explanation:

The pressure is constant, so we can use Charles' Law to calculate the volume.

$\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}$

Data:

V₁ = 693 mL; T₁ = 45 °C

V₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert temperature to kelvins

T₁ = ( 45 + 273.15) = 318.15 K

T₂ = (150 + 273.15) = 423.15 K

(b) Calculate the volume

$\dfrac{ 693}{318.15} = \dfrac{ V_{2}}{423.15}\\\\2.178 = \dfrac{ V_{2}}{423.15}\\\\V_{2} = 2.178 \times 423.15 = \boxed{\textbf{922 mL}}$

5 0

2 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

2 years ago