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11Alexandr11 [23.1K]
3 years ago
9

You take 50 mL of small BB's and combine them with 50 mL of large BB's and you get a total of 90 mL BB's of mixed size. Explain

Chemistry
1 answer:
Snezhnost [94]3 years ago
7 0
The 50 mL of large BB's is actually less than the 50 mL of small BB's because of size difference
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Which electron configuration of the 4f energy sublevel is most stable?
umka21 [38]

Answer: both options A and D

Explanation:

Half filled and completely filled orbitals are more stable than any other configuration since they are more symmetrical and energy exchange Occurs readily.

So 4f^7& 4f^14 are more stable

7 0
3 years ago
What is the name of the compound H₃S₅?
Usimov [2.4K]
The compound name for H3S5 is hydrosulfide sulfanide sulfide
7 0
2 years ago
How many atoms are in 578g Na
borishaifa [10]

1.51 x 10²⁵atoms

Explanation:

Given parameters:

Mass of Na = 578g

Unknown:

Number of atoms = ?

Solution:

 To find the number of atoms, we must first find the number of moles the given mass contains.

  Number of moles  = \frac{mass}{molar mass}

   molar mass of Na = 23g

 Number of moles =  \frac{578}{23} =  25.13moles

   1 mole of a substance = 6.02 x 10²³atoms

    25.13 mole of Na = 25.13 x  6.02 x 10²³atoms

 This gives  1.51 x 10²⁵atoms of Na

Learn more:

Avogadro's constant brainly.com/question/2746374

#learnwithBrainly

5 0
3 years ago
(a) a 0.2 m potassium hydroxide solution is titrated with a 0.1 m nitric acid solution. (i) balanced equation: (ii)what would be
Stells [14]

\text{KOH} (aq) + \text{HNO}_3 (aq) \to \text{KNO}_3 (aq) + \text{H}_2\text{O} (l)

The solution shall contain only \text{KNO}_3 (aq) (and water) at the equivalence point. Both potassium hydroxide and nitric acid exist as strong electrolytes. As a result,  \text{KNO}_3 (aq), the salt derived from a reaction between the two species would undergo hydrolysis of a negligible extent. This neutralization reaction therefore be neutral at the equilibrium point.

The question states that the solution is "titrated with a ... nitric acid solution" indicating that \text{HNO}_3 is added to the initially-basic solution. PH value of the solution would keep decreasing as the volume of the acid added increases. The final solution would be acidic as it contains not only water and \text{KNO}_3 (aq), but some \text{HNO}_3 as well. Bromothymol blue would therefore demonstrates a yellow color, the color it present in an acidic solution, at the end of the titration.

5 0
3 years ago
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0
3 years ago
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