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Which element would you expect to be more metallic?<br> (a) Ca or Rb (b) Mg or Ra (c) Br or I

svlad2 [7]

Explanation:

When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.

As calcium (Ca) is a group 2A element and rubidium (Rb) is a group 1A element. Hence, Rb being an alkali metal is more metallic in nature than calcium (alkaline earth metal).

Both magnesium (Mg) and radium (Ra) are group 2A elements. And, when we move down a group then as the size of element increases so, it becomes easy of the metal atom to lose an electron.

As a result, there occurs an increase in metallic character of the element. Hence, Radium (Ra) is more metallic in nature than magnesium (Mg).

Also, both bromine and iodine are group 17 elements. Since, both of them are non-metals and non-metallic character increases on moving down the group.

Therefore, bromine (Br) is more metallic than iodine.

7 0

3 years ago

A partir de la desintegración del Virreinato del Río de la Plata y durante todo el siglo XXI, los nuevos Estados nacionales que

nalin [4]

Answer:

Verdadero (True).

Explanation:

Después de la desintegración del citado virreinato, los nuevos Estados intentaron establecer sus fronteras, a menudo a través de guerras e invasiones. Cabe destacar la invasión brasileña de Uruguay o la Guerra de la Triple Alianza. En menor medida, lo hicieron a través de tratados internacionales. (After the disolution of the viceroyalship described above, the new States attempted to establish their frontier usually by wars and invasions. It is to highlight the Brazilian invasion of Uruguay or the Triple Alliance' War. In a lesser extent, they made it through international treaties.)

3 0

3 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

Dan bikes 10 km west and then bikes another 5 km west. What is Dan's

777dan777 [17]

C

This is because 10+5=15
15/45=0.3

4 0

3 years ago

you have a solution of water and salt. if the total solution weighs 50 grams and the salt weighs 4 grams. what is the percent co

Natali5045456 [20]

Answer:

92%

Explanation:

I believe. Hope this was helpful.

8 0

3 years ago