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Mrrafil [7]
1 year ago
6

At room temperature (20°C) and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

n that of air. In a buoyancy experiment with a new plastic, a chemist creates a rigid, thin-walled ball that weighs 0.12 g and has a volume of 560 cm⁹.
(c) Will it float if filled with hydrogen (d = 0.0899 g/L)?
Chemistry
1 answer:
Eduardwww [97]1 year ago
8 0

Total density of filled ball with hydrogen gas: \frac{0.12g+0.050344g}{0.560L} = 0.3041g/L

The relationship between mass and volume can be easily determined using density; for example, the mass of a body is equal to its volume multiplied by the density (M = Vd), whereas the volume is equal to the mass divided by the density (V = M/d). The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

Density of hydrogen gas = d_2 = 0.0899g/L

Mass of the hydrogen gas : \frac{0.12g+0.050344}{0.560L} = 0.3041g/L

Learn more about Mass and Density here:

brainly.com/question/10821730

#SPJ4

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Please Help !! <br><br> The weak base ionization<br><br> constant (Kb) for CIO is<br><br> equal to:
Zolol [24]

Answer:

k it I did I'd help ya

Explanation:

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3 0
2 years ago
Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr
Ganezh [65]

Answer:

a. 4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas N_2, oxygen gas O_2, water vapor H_2O and carbon dioxide CO_2. Let's write the decomposition of nitroglycerin into these 4 components:

C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by \frac{3}{2}:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by \frac{5}{2}:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

\frac{5}{2} + 6 = 8.5

This leaves 9 - 8.5 = 0.5 = \frac{1}{2} of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)

To make it look neater without fractional coefficients, multiply both sides by 4:

4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

V_{CO_2} = 41.0 L

T = -14.0^oC + 273.15 K = 259.15 K

p = 1 atm

Firstly, we may find moles of carbon dioxide produced using the ideal gas law pV = nRT.

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol

According to the stoichiometry of the balanced chemical equation:

4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g

3 0
3 years ago
Given the following reaction: NH4SH (s) &lt;--&gt; NH3 (g) + H2S (g) If we start
almond37 [142]

Answer:

D. 0.3 M

Explanation:

                                              NH4SH (s)      <-->            NH3 (g) + H2S (g)

Initial concentration              0.085mol/0.25L             0                 0

Change in concentration     -0.2M                               +0.2 M        +0.2M

Equilibrium               0.035mol/0.25 L=0.14M             0.2M           0.2M

concentration

Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M

K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M

4 0
3 years ago
When CO2 and H2O react to form H2CO3, the number of bonds broken in H2O is ____ and in CO2 is _____. one; one
guapka [62]

CO2 and H2O react to form H2CO3 and two bonds are broken each in CO and H2O to form H2CO3.

<h3>What is chemical bonding?</h3>

Chemical bonding refers to the forces of attraction which hold atoms of the same or different elements together in order to form stable compounds or molecules .

Chemical bonding may be either ionic or covalent.

The greater the number of bonds in a compound, the more stable the compound.

During chemical reactions, bonds are broken and new binds are formed.

There are two bonds each in CO2 and H2O.

This, in the reaction between CO2 and H2O react to form H2CO3, , the number of bonds broken in H2O is two and in CO2 is two.

Learn more about chemical bonding at: brainly.com/question/819068

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2 years ago
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