All categories

3 years ago

If acids are compounds that release protons (h+), how is it that sox and nox cause acid rain?

1 answer:

3 0

The protons come from water. Sulfur oxides dissolve in water to form sulfurous and sulfuric acids, and nitrogen oxides dissolve to form nitrous and nitric acids.

You might be interested in

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

Mr. McPhillip has a ribbon of magnesium (Mg), a silvery metal. As a demonstration for his class, he sets the metal on fire. It b

OLEGan [10]

A. Magnesium is an element and after it is burned it is a compound

5 0

3 years ago

Read 2 more answers

Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)

cupoosta [38]

Answer:

$m_{CaCO_3}=0.179gCaCO_3$

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2$

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

$m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3$

Regards.

3 0

4 years ago

What answer giving brainliest tysm

monitta

Answer:

Explanation:

it is repeating

3 0

3 years ago

5. The Ks for silver chloride is 1.8 x 10-lat

AnnZ [28]

Answer:

bueno

Explanation:

hijonshsjijagha

6 0

3 years ago