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Alexus [3.1K]
3 years ago
9

Period, block, and group

Chemistry
1 answer:
Sedaia [141]3 years ago
8 0

Answer:

what do you mean

Explanation:

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Matter is every object that stands on this Earth
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3 years ago
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predict where an unknown element that has these properties could fit in the periodic table 62 protons and electrons
Vesnalui [34]
Position of element in periodic table is depend on the electronic configuration of element.

Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.

Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.

Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table. 
5 0
3 years ago
Butane (C4H10) is used as a fuel where natural gas is not available. How many grams of butane will fill a 3.50-liter container a
erik [133]
<span>We use the formula PV = nRT. P = 758 torr = 0.997 atm. V = 3.50 L. T = 35.6 C = 308.15 K. R = 0.0821. Rearranging the equation gives up n = PV/Rt and we get .0138 moles of butane. Mass of 0.0138 moles of butane = .0138 x 58.12 = 8.02g.</span>
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3 years ago
How do i design a controlled experiment to appropriately test a hypothesis?
Virty [35]

Answer:

you need only one independent variable because if not, you wont know what factors have changed your experiment.

Explanation:

3 0
3 years ago
How much heat energy would be released if 78.1 g of water at 0.00 °c were converted to ice at −57.1 °c. give your answer as a po
kolezko [41]
To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g; 
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C 
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348 
 = 26250.0348 J or 26.250 kJ
3 0
3 years ago
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