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4 years ago

What would you use if you needed to determine the density of an object

1 answer:

8 0

In order to find the density of an object, you need to know the mass of the object and the volume.
That is why the unit of density is kg/m3.
Hope it helps

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If element X has 83 protons, how many electrons does it have? ______ electrons

Zolol [24]

The element has 83 electrons

5 0

3 years ago

Why is it impossible to have an undiscovered element that would fit between sodium and magnesuim

Stolb23 [73]

Answer:

Sodium has 11 protons and Magnesium has 12. The way that the periodic table is set up doesn't allow for any elements to be in between. The element would have to have 11.5 protons, which is impossible

4 0

3 years ago

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

riadik2000 [5.3K]

Answer:

A) SiO2 is the limiting reactant

B) Theoretical yield= 72333.3g

C) % yield =91.5%

Explanation:

SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)

n(SiO2)= 155000/60 = 2583.33 mols

n(C)= 79000/12= 3291.66 mols

a)SiO2 is the limiting reactant

According to the balanced reaction equation,

60g of SiO2 produced 28g of SiO2

155000g of SiO2 will produce 155000×28/60= 72333.3g

Therefore theoretical yield of Si= 72333.3g

% yield= 66200/72333.3×100/1 =91.5%

5 0

3 years ago

Please help It’s super quick

UkoKoshka [18]

Answer:

1. a

Explanation:

2. c

7 0

3 years ago

Read 2 more answers

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:

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