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2 years ago

What type of precipitation occurs when the temperature of the sky is above freezing? Below

Chemistry

2 answers:

Shkiper50 [21]2 years ago

6 0

Answer:

Above freezing :- rain

Below freezing :- hail

Damm [24]2 years ago

6 0

Answer:

below freezing

Explanation:

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They dump large amounts of rainwater that causes flooding. *

IgorC [24]

Answer =Tsunamis because they flood places and areas

3 0

2 years ago

Read 2 more answers

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

Which of the following represents the least number of molecues?

Mars2501 [29]

Answer:

Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23

Explanation:

In order to find the amount of molecules of each of the options, we need to follow the following equation.

$molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}$

So, let´s get the number of molecules for each of the options.

$a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules$

$b) molecules=\frac{77(g)x6.022x10^{23}(molecules/mol) }{16.06(g/mol)}=2.89x10^{24}molecules$

$c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules$

$d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules$

$d) molecules=\frac{84(g)x6.022x10^{23}(molecules/mol) }{20.01(g/mol)}=2.53x10^{24}molecules$

the smalest number is in option a)

Best of luck.

7 0

3 years ago

A solution containing cacl2 is mixed with a solution of li2c2o4 to form a solution that is 2. 1*10^-5m in calcium ion and 4. 75*

vladimir1956 [14]

Answer:

the answer is d

Explanation:

i know

4 0

2 years ago

Using the atomic masses and relative abundance of the isotopes of nitrogen given below, determine the average atomic mass of nit

mylen [45]

The formula for average atomic mass is :

mass of isotope A * % of isotope A + mass of isotope B * % of isotope B + ....

Now,
Here,
Average atomic mass of nitrogen = (14.003 * 99.63%) + (15.000 * 0.37%)
= (14.003 * 0.9963) + ( 15.000 * 0.0037)
= 13.951 + 0.056
= 14.007 a.m.u.

3 0

3 years ago