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kirill115 [55]
2 years ago
15

How many moles of O are in 11.6 mol Fe(NO3)3?

Chemistry
1 answer:
Westkost [7]2 years ago
6 0

Answer:

Why might the expression "1 mole of nitrogen" be confusing? ... WebAssign will check your answer for the correct number of significant ... Calculate the moles of O atoms in 8.18 g of Mg(NO3)2. ... If the molar mass of hemoglobin is 6.8 104 g/mol, how many Fe2+ ions are ... (_) C7H16(l) + 11 O2(g) 7 CO2(g) + 8 H2O(g)

Explanation:

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lava is coming from the crust

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volcano's explode with lava because the crust pushes it out

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2 years ago
Give two different conditions under which long chain hydrocarbons may be cracked.​
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In thermal cracking, high temperatures (typically in the range of 450°C to 750°C) and pressures (up to about 70 atmospheres) are used to break the large hydrocarbons into smaller ones. Thermal cracking gives mixtures of products containing high proportions of hydrocarbons with double bonds - alkenes.
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3 years ago
Please help... I am begging...
iVinArrow [24]

Answer:

The number, such as 98.7 FM, of a radio station represents:

  • <u>the frequency in which is transmitted the radio signal</u>.

Explanation:

<em>The radio FM is the modulated frequency, which means that all the information is sent by just a signal, with different frequencies which difference them</em>, <u>the radio FM use the frequencies from 88 MHz until 108 Mhz (MHz is a measuring unit for the frequency), with a minimal space among them of 0.2 MHz</u>, this last means that you could find a signal in 88.0 MHz, and the next should be 88.2 MHz, next 88.4 MHz and so (at least, regularly the space between two frequencies is more than 0.2 MHz).

5 0
3 years ago
How many grams of calcium phosphate can be produced when 78.5 grams of calcium hydroxide reacts with excess phosphoric acid? Unb
sasho [114]
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3 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]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

  • <u>For a:</u>

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

<u>Conversion factor:</u>  1cm=10^{10}pm

Hence, the edge length of the crystal is 533.5 pm

  • <u>For b:</u>

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
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