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allsm [11]
3 years ago
14

Calculate the density of an object with a mass of 30g and volume of 6cm^3

Chemistry
1 answer:
Kipish [7]3 years ago
3 0
Density = Mass ÷ Volume

D= 30g ÷ 6 cm^3

D= 5 g/dm^3

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6 0
3 years ago
Explain why both square planar and tetrahedral complexes have coordination number=4, and yet square planar complexes can never b
saul85 [17]

Answer:

The coordination number is 4.  

Explanation:

  • Square planar clusters can be either cis or trans, as they form 180 and 90-degree bond angles. Therefore, a pair of ions may be adjacent (cis) to one another and immediately across (trans) from one another.
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7 0
3 years ago
Calculate the volume in liters of a ×1.0310−6mM silver(II) oxide solution that contains 900.mg of silver(II) oxide AgO.
sesenic [268]

Answer : The volume of solution will be, 7.047\times 10^6L

Explanation : Given,

Mass of AgO = 900 mg  = 900000 g

conversion used : (1 mg = 1000 g)

Molar mass of AgO = 124 g/mole

Molarity of AgO = 1.03\times 10^{-6}mM=0.00103M

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{volume of solution in liter}}

Now put all the given values in this formula, we get the active mass of urea.

0.00103M=\frac{900000g}{124g/mole\times \text{volume of solution in liter}}

\text{volume of solution in liter}=7046664.579L=7.047\times 10^6L

Therefore, the volume of solution will be, 7.047\times 10^6L

6 0
3 years ago
Explain the units and how to solve the following metric conversion then solve the metric conversion by filling in the blank 12KM
LiRa [457]

Answer:- 12 km = 12000 m

Solution:- It's a metric unit conversion where we are asked to convert 12 km to m where km stands for kilometer and m stands for meter.

In metric conversions, kilo means 1000.

So, 1 km = 1000 m

It means,  we multiply the given km by 1000 to get the answer in m as:

12km(\frac{1000m}{1km})

= 12000 m

Hence, 12 km = 12000 m.

5 0
4 years ago
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