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zaharov [31]
3 years ago
13

Give the maximum number of electrons in an atom that can have these quantum numbers:\

Chemistry
1 answer:
jeka57 [31]3 years ago
4 0
Please add a picture of what numbers ; )
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1. A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C. If the
Maurinko [17]

Answer:

Mass of aluminium in sample = 3.591 g ≅ 3.6 grams

Explanation:

Given that,  A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.

the specific heat of aluminum is 0.900 J/g- °C

The relation between heat absorbed and change in temperature is given by,   Q = msΔT.

where Q = heat absorbed

            m = mass of the substance

            s = specific heat of substance

          ΔT  = change in temperature

Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C

⇒ m =  \frac{Q}{s(T_{2} -T_{1}) }

⇒ m = \frac{50.1}{0.900(15.5)} = 3.591 g ≅ 3.6 g

⇒ m ≅ 3.6 g

5 0
3 years ago
The heat of vaporization of water is 540 cal/g. How much heat is needed to change 4.28g of water to steam?
damaskus [11]

Answer:

2311.2 cal

Explanation:

540 cal / g    *   4.28 g = 2311.2  cal

7 0
2 years ago
What's the lowest to greatest order for scales of universe?
elena55 [62]

Answer:

universe

Explanation:

tell me if im wrong

4 0
3 years ago
Consider a 175.7 g sample of the compound manganese(IV) perchlorate.
kramer

Answer:

6.21 moles O

Explanation:

To find the moles of oxygen, you need to (1) convert grams Mn(ClO₄)₄ to moles Mn(ClO₄)₄ (via molar mass) and then (2) convert moles Mn(ClO₄)₄ to moles O (via mole-to-mole ratio from formula subscripts). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.

Molar Mass (Mn(ClO₄)₄): 452.74 g/mol

1 Mn(ClO₄)₄ = 1 Mn and 4 Cl and 16 O

175.7 g Mn(ClO₄)₄           1 mole                   16 moles O
---------------------------  x  -------------------  x  ---------------------------  =  6.21 moles O
                                       452.74 g            1 mole Mn(ClO₄)₄

6 0
2 years ago
9.69×10^25 formula units of iron(III) nitrate is equal to how many moles of Fe(NO3)3?
scoray [572]

Answer:

160.9 mol ≅ 161.0 mol.

Explanation:

  • It is known that every 1.0 mole of compound or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms (formula units).

Using cross multiplication:

1.0 mole of Fe(NO₃)₃ contains → 6.022 x 10²³ formula units.

??? mole of Fe(NO₃)₃ contains → 9.69 x 10²⁵ formula units.

<em>∴ The no. of moles of He contains (9.69 x 10²⁵ formula units)</em> = (1.0 mol)(9.69 x 10²⁵ formula units.)/(6.022 x 10²³ formula units) = <em>160.9 mol ≅ 161.0 mol.</em>

6 0
3 years ago
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