Sodium(na) is a highly reactive metal and chlorine is a poisonous gas. Which compound do they form when they chemically combine?

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12

Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

$120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu$

7 0

3 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

6 0

3 years ago

Which of these statements best explains why the soil around a volcanic region is fertile?

AVprozaik [17]

Answer:

The correct statement is option c, that is, particles discharged in the air by volcanoes fall to the ground and enrich the soil.

Explanation:

The eruptions of volcanoes lead to the dispersion of ash over the broader regions surrounding the site of eruption. On the basis of the chemistry of the magma, the ash will be comprising different concentrations of soil nutrients. While the major elements found in the magma are oxygen and silica, the eruptions also lead to the discharging of carbon dioxide, water, hydrogen sulfide, sulfur dioxide, and hydrogen chloride.

In supplementation, the eruptions also discharge bits of rocks like pyroxene, potolivine, amphibole, feldspar that are in turn enriched with magnesium, iron, and potassium. As an outcome, the areas which comprise huge deposits of the volcanic soil are quite fertile.

3 0

4 years ago

Read 2 more answers

How would your atomic mass prediction have been affected if not all of the water had been evaporated from the calcium carbonate

Yanka [14]

Go to a famlie member, if you can't trust them, trust god. if you cant trust god, BURN IN HELL

6 0

3 years ago