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3 years ago

What is the oxidation number of HCLO4?

1 answer:

4 0

E.g. in H3PO4 (O, -2).
8. The sum of the oxidation states of all the atoms in a species must be equal to net charge on the species. e.g. Net Charge of HClO4 = 0, i.e. [+1(H)+7(Cl)-2*4(O)] = 0.

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The different phases of the Moon refer to A. Changes in the Moon's shape. B. Changes in the Moon's visibility. C. Changes in sea

Sav [38]

Answer:

B. Changes in the Moon's visibility.

Explanation:

The moon's shifting of position makes it "appear" as being smaller or bigger. This is simply because the moon is rotating and the sun's light shines on a certain area of the moon, and depending on where we are on the Earth is how the moon looks or appears to our perspective.

5 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

Answer:

For a: The edge length of the crystal is 533.5 pm

For b: The atomic radius of potassium is 231.01 pm

Explanation:

For a:

To calculate the lattice parameter or edge length of the crystal, we use the equation:

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

where,

$\rho$ = density = $0.855g/cm^3$

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

M = atomic mass of metal = 39.09 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:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

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

Hence, the edge length of the crystal is 533.5 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 = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

4 0

3 years ago

A student titrates a solution containing 0.050 moles of calcium hydroxide with phosphoric acid. How many moles of phosphoric aci

lina2011 [118]

Answer:

The answer to your question is 0.33 moles of H₃PO₄

Explanation:

Data

moles of Ca(OH)₂ = 0.050

moles of H₃PO₄ = ?

Process

1.- Write the balanced chemical equation

3Ca(OH)₂ + 2H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6H₂O

Reactants Elements Products

3 Ca 3

12 H 12

14 O 14

2.- Calculate the moles of phosphoric acid

3 moles of calcium hydroxide --------------- 2 moles of phosphoric acid

0.5 moles of calcium hydroxide ----------- x

x = (0.5 x 2)/3

x = 0.33 moles

6 0

3 years ago

A gold cube is 170.00 mm long, 20.00 cm wide, and 0.99 m thick. If gold has a

GaryK [48]

Answer:

Explanation:

There's 2 steps: first find the volume then the mass

volume=lwh

convert 20 cm to mm: 20x10=200mm

volume=(170)(200)(0.99)

volume=33,660 mm

since the density is in cm cubed, convert 33,660 mm to cm

33,660/10=3366

density=mass/volume, rearrange to get mass=density x volume

mass=19.3 x 3366

mass=64963.8

6 0

3 years ago

Imagine that you burn enough natural gas to release one giga-joule of energy (109 J). Estimate the mass of CO2 released into the

sweet [91]

Answer:

5.0 × 10⁴ g of CO₂

Explanation:

First, we write the balanced equation for the combustion of methane.

CH₄ + 1.5 O₂ ⇄ CO₂ + H₂O

Now, we can establish the following relations:

16g of CH₄ produce 44g of CO₂
1 GJ = 10⁹ J
55 MJ (55 × 10⁶ J) are released for every 1 kg (10³g) of CH₄ burned.

Using proportions,

$10^{9} J.\frac{10^{3}gCH_{4} }{55 \times 10^{6} J } .\frac{44gCO_{2}}{16gCH_{4}} =5.0 \times 10^{4} g CO_{2}$

6 0

3 years ago