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3 years ago

Can you please help me with the second one please can you please show your work please

Chemistry

1 answer:

Alex787 [66]3 years ago

8 0

Answer:

10800 seconds

Explanation:

3 x 60 = 180

180 x 60 = 10800

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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

4 years ago

7. Convert 5.2 cm of magnesium (Mg) ribbon to mm of Mg ribbon.

Effectus [21]

<h3>Answer:</h3>

52 mm

<h3>Explanation:</h3>

We are given;

5.2 cm of magnesium

Required to convert it to cm

We are going to use the appropriate conversion factor;

The units used to measure length include;

Kilometer(km)

Hectometer (Hm)

Decameter (dkm)

Meter(m)

Decimeter (dm)

Centimeter (cm)

Millimeter (mm)

Therefore; the appropriate conversion factor is 10mm/cm

Thus;

5.2 cm will be equivalent to;

= 5.2 cm × 10 mm/cm

= 52 mm

Therefore, the length of magnesium ribbon is 52 mm

7 0

3 years ago

Read 2 more answers

A thermometer is placed in a beaker full of water that has been sitting on the lab table for a few hours. The beaker is placed o

krek1111 [17]

Based on the information I would assume B, 73 degrees...

It shouldn't be A, 4 minutes on the burner should increase the temperature.

If it were D, it would be beyond boiling, and water takes a decent amount of energy to heat, D should be all vapor.

Same logic for C, it's basically almost boiling.

I would say 73 degrees seems most reasonable for 4 minutes.

7 0

4 years ago

Read 2 more answers

Question 1 How many grams are equal to 3.4 moles of carbon dioxide?

ddd [48]

Answer:

149.6 grams

Explanation:

Mass in gram = molar mass * number of moles

Massof CO2 in gram = 44*3.4=149.6 grams

4 0

3 years ago

If E=mc^2,solve for both m and c. Also, if m=80 and c=0.40 what is the value of E?

Pie

Answer:

$m = \frac{E}{c^2}$

$c = \sqrt{\frac{E}{m} }$

$E = 12.8 J$

Explanation:

Part 1: Solving for m

We are given that:

E = mc²

To solve for m, we will need to isolate the m on one side of the equation

This means that we will simply divide both sides by c²

$m = \frac{E}{c^2}$

Part 2: Solving for c

We are given that:

E = mc²

To solve for c, we will need to isolate the m on one side of the equation

This means that first we will divide both sides by m and then take square root for both sides to get the value of c

$c^2 = \frac{E}{m}\\ \\ c=\sqrt{\frac{E}{m}}$

Part 3: Solving for E

We are given that:

m = 80 and c = 0.4

To get the value of E, we will simply substitute in the given equation:

E = mc²

E = (80) × (0.4)²

E = 12.8 J

Hope this helps :)

4 0

3 years ago