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ale4655 [162]
3 years ago
7

A box is 1 m high, 2.5 m long, and 1.5 m wide, its volume is 5 mº. true or false

Chemistry
2 answers:
elena-s [515]3 years ago
7 0

Answer:

False

Correct answer: 3.75m³

Explanation:

Given parameters:

  Height of box  = 1m

  Length of box  = 2.5m

  Width of box  = 1.5m

Unknown:

Volume of the box  = ?

Solution:

The volume of the box is given as:

   Volume  = length  x height x width

 Volume  = 2.5 x 1 x 1.5

 Volume  = 3.75m³

Maurinko [17]3 years ago
5 0

Answer:

false

Explanation:

Because volume is in m3

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4. How many milligrams are in 5.25 x 10-13 kg?<br><br> the “-13” is an exponent
rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

   Given mass = 5.25 x 10⁻¹³kg

The kilo-  is a prefix that denotes 10³

  therefore;

         1000g = 1kilogram

 the milli-  is a prefix that denotes 10⁻⁻³

       1000mg = 1g

Now that we know this, we can convert:

   5.25 x 10⁻¹³kg  x \frac{1000g}{1kg}  x \frac{1000mg}{1g}   =  5. 25 x 10⁻¹³ x 10⁶mg

      =  5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0
3 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
Define organic chemistry and inorganic chemistry.
MrRissso [65]
Organic is safer inorganic is the same but less better
3 0
3 years ago
Read 2 more answers
A) Atom is generally charge less,<br>why? ​
Aliun [14]

Answer:

Explanation:

Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons. These opposite charges cancel each other out making the atom neutral.

7 0
3 years ago
What will the pressure be if 89.9 moles of argon are contained in a 12.0 L cylinder that is pressurized at a temperature of 300
irinina [24]
  • P=nRT/V
  • p=89.9(8.314)(12)/300
  • P=8969.14/300
  • P=29.89atm
  • P=29.9atm

Done!

7 0
2 years ago
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