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Crank
2 years ago
10

Which of these elements has the smallest atomic radius?

Chemistry
1 answer:
nadya68 [22]2 years ago
7 0

Answer:

3. be

Explanation:

it shows it in the word search

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Consider the following reaction: C6H6 + O2 \longrightarrow ⟶ CO2 + H2O 39.7 grams of C6H6 are allowed to react with 105.7 g of O
Ivenika [448]

Answer:

116.3 grCO2

Explanation:

1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side

C6H6 +15/2 O2⟶ 6CO2 +3 H2O

2nd - we calculate the limiting reagent

39.2gr C6H6*(240grO2/78grC6H6)=120 grO2

we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent

3rd - we use the limiting reagent to calculate the amount of CO2 in grams

105.7grO2*(264grCO2/240grO2)=116.3 grCO2

7 0
2 years ago
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]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

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

  • <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 = 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
Lithium (Li) and oxygen (O) are both in period 2. They both have? a.similar properties
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They both have two electron shells. Period indicates number of shells.
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The charge of a proton is positive.
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The molecular formula of glucose:
C₆H₁₂O₆
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