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valina [46]
4 years ago
12

How many moles are in 156 grams of potassium??? I already tried to ask google : (

Chemistry
2 answers:
Y_Kistochka [10]4 years ago
5 0

Answer:

1 mole is equal to 1 moles Potassium, or 39.0983 grams.

39.0983g X 4 = 156.3932

so 4 moles are in 156 grams of Potassium

Explanation:

Hope this helps

-A Helping Friend (mark brainliest pls)

Fofino [41]4 years ago
5 0
Mole =Mass(g)
———
RMM(relative molecular mass)(g/mol)
N.B The RMM is the same as the mass number but should not be used interchangeably because the mass number has no units while the RMM does which g/mol

Mole =156/39
= 4 mol
Mark me as the brainliest❤️✌
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Zinc phosphide, Zn3P2, is often used as a rat poison. Phosphorus has 5 valence electrons. Assuming that phosphorus is following
Nadya [2.5K]

Answer:

Two electrons each!

Explanation:

The question pretty much requires us to find the oxidation number of Zinc in the compound.

Zn3P2

Following oxidation number rules;

O.N of Zn3P2 = 0

.Since phosphorus has valence of 5, it needs three more electrons to achieve its octet state. Hence;

Oxidation number of P = -3

Let oxidation number of Zn = x

We have;

3x  +  2 (-3) = 0

3x + (-6) = 0

3x = 6

x = 6/3 = 2

This means each zn electrons lost 2 electrons.

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3 years ago
What is the Net Force?<br> 2N &gt;<br> 4N -<br> SN
maria [59]
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Hot lead with a mass of 200.0 g of (Specific heat of Pb = 0.129 J/gºC) at 176.4°C was dropped into a calorimeter containing an u
Morgarella [4.7K]

the Calorimetry relationships you can find the amount of water in the calorimeter is      m = 21.3 g

given parameters

  • Lead mass M = 200.0 g
  • Initial lead temperature T₁ = 176.4ºC
  • Specific heat of Lead    c_{e Pb} = 0.129 J / g ºC
  • Sspecific heat of water c_{e H_2O} = 4.186 J / g ºC
  • Initial water temperature T₀ = 21.7ºC
  • Equilibrium temperature T_f = 56.4ºC

to find

The body of water

Thermal energy is the energy stored in the body that can be transferred as heat when two or more bodies are in contact. Calorimetry is a technique where the energy is transferred between the body only in the form of heat and in this case the thermal energy of the lead is transferred to the calorimeter that reaches the equilibrium that the thematic energy of the two is equal

              Q_{ceded} = Q_{absorbed}

               

Lead, because it is hotter, gives up energy

              Q_{ceded} = M c_{e Pb} (T₁ - T_f)

The calorimeter that is colder absorbs the heat

              Q_{absrobed} = m c_{e H_2O} (T_f - T₀)

where M and m are the mass of lead and water, respectively, c are the specific heats, T₁ is the temperature of the hot lead, T₀ the temperature of cold water and T_f the equilibrium temperature

              M c_{ePb} (T₁ - T_f) = m c_{eH2O} (T_f - T₀)

               m = \frac{ M\  c_{ePb} \ (T_1 - T_f)}{c_{eH_2O}  \ (T_f - T_o)}

let's calculate

              m = \frac{200 \ 0.129  (176.4-56.4)}{ 4.186 \  (56.4 -21.7)}

              m = 3096 / 145.25

              m = 21.3 g

Using the Calorimetry relationships you can find the amount of water in the calorimeter is:

             m = 21.3 g

learn more about calorimetry here:

brainly.com/question/15073428

6 0
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2. Oxygen because it has more mass
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