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White raven [17]
3 years ago
10

hydrogen and helium are different elements. how can you use the plum pudding model to show how atoms of the two elements might b

e different from one another
Chemistry
1 answer:
lesya [120]3 years ago
6 0

Answer:

Hydrogen and helium are different elements. How can you use the plum pudding model to show how atoms of the two elements might be different from one another? Answer: Possible answer: The two types of atoms could have different amounts of positive fluid and different numbers of electrons.

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How much Cu-64 (half-life about 12 hours) would remain from a 2 mg sample after 12 hours?​
Bogdan [553]

Answer:

i beleive it is  A if not here a cookie online cookie also dont forget to wear your mask kiddos

Explanation:

7 0
3 years ago
Read 2 more answers
O, li, c which electrons experience the greatest effective nuclear charge? express your answer as a chemical formula.
Nataly_w [17]
<span>The effective nuclear charge of an atom = total electrons - inner electrons For O, ENC = 8 - 2 = 6 For Li, ENC = 3 - 2 = 1 For C, ENC = 6 - 2 = 4 The electrons in O experience the greatest effective nuclear charge and that is why O is smaller than C (which is smaller than Li).</span>
5 0
4 years ago
Write a balanced equation for zinc sulfate heptahydrate + sodium carbonate —&gt; zinc carbonate + sodium sulfate
tensa zangetsu [6.8K]

Answer:

I gotta go - but think heat or solubility, what can you drive away with heat and by what is it is not soluble...it is a catch all answer - but rarely is there ONE WAY ONLY! Good luck.

Explanation:

3 0
3 years ago
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho
timama [110]

Explanation:

As per the given data, at a higher temperature, at 24^{o}C, the solution will occupy a larger volume than at 20^{o}C.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

                 = \frac{C_{1}}{d_{1}}

                 = \frac{1.000 mol/L}{0.9982071 g/ml}

                 = 1.0017961 mol/g  

At 24^{o}C, density of water = 0.9972995 g/ml

Since, \frac{concentration}{density} = \frac{C_{2}}{d_{2}}

                           = \frac{C_{2}}{0.9972995}

Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

so,                   1.0017961 mol/g = \frac{C_{2}}{0.9972995}

                      C_{2} = 1.0017961 \times 0.9972995

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of KNO_{3} present is calculated as follows.

             C_{2} = \frac{concentration}{volume}

               0.9990907 mol/L = \frac{concentration}{0.5 L}  

               concentration = 0.49954537 mol

Hence, mass (m'') = 0.49954537 mol \times 101.1032 g/mol = 50.5056 g       (as molar mass of KNO_{3} = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}

where,          d_{air} = density of air = 0.0012 g/ml

                     d_{weight} = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = 50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of KNO_{3} needs to be measured.

8 0
3 years ago
Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is 7.60 feet deep, 100.0 yards wi
Effectus [21]

Answer:

[MCHM] = 7.52 M

Explanation:

This all about unit conversion

1 feet = 0.3048 meters

1 yard = 0.9144 meters

7.60 feet . 0.3048 = 2.31 meters deep

100 yd . 0.9144 meters = 91.44 meters long and 91.44 meters wide

In the river we have a volume of:

2.31 m . 91.44m . 91.44m = 19314.5  m³

1 dm³ = 1 L

1 dm³ = 0.001 m³

19314.5 m³ / 0.001 m³ = 19314542 L (The final volume of the river)

3.785 L = 1 gallon

In 19314542 L, we have (19314542 / 3.785) = 5102917 gallons

1 gallon = 128 oz

5102917 gallons . 128 = 653173376 oz

1 oz = 28.3495 g

653173376 oz . 28.495 = 1.86x10¹⁰ grams

Molar mass MCHM = 128 g/m

Moles of MCHM = 1.86x10¹⁰ grams / 128 g/m = 145312500 moles

Molarity = mol /L → 145312500 moles / 19314542 L = 7.52 M

7 0
4 years ago
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