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Viktor [21]
3 years ago
8

The electron configuration 1s2 is correct for an element that has 2 electrons, both in the first principle energy level

Chemistry
1 answer:
Molodets [167]3 years ago
5 0

Answer:

The answer is True

Explanation:

I got it right on my quiz, hope this helps!

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How many molecules of glucose are in 5. 72 grams of glucose c6h12o6?.
liq [111]

Answer:

5.72 g ( 1 mol / 180.16 g ) (  6.022 x 10^23 molecules / mole ) = 1.90x10^23 molecules

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2 years ago
COMMONLIT: WATER SCARCITY: A GLOBAL ISSUE
aalyn [17]
Plz help me I need help to but in math plzz I have 2 miplz help me
7 0
3 years ago
Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla
Pavel [41]

Answer:

ΔH = +26.08 kJ/mol

Explanation:

The change in enthalpy (ΔH) is given in J/mol, and can be calculated for  dissolution by the equation:

ΔH = m(water)*Cp*ΔT/n(solute)

The mass of water is the density multiplied by the volume

m = 1g/mL * 25.0mL = 25.0 g

The number of the moles is the mass divided by the molar mass. Knowing the molar masses of the elements:

N = 14 g/mol x 2 = 28

H = 1 g/mol x 4 = 4

O = 16 g/mol x 3 = 48

NH₄NO₃ = 80 g/mol

n = 1.25/80 = 0.015625 mol

So,  

ΔH = 25*4.18*(25.8 - 21.9)/0.015625

ΔH = 26,083.2 J/mol

ΔH = +26.08 kJ/mol

5 0
3 years ago
If a boy eats a meal and then runs
aalyn [17]

Answer: A

Explanation: A, because when you eat it is you use chemical energy afterward you run and use mechanical energy. Ex of Chemical energy can be batteries and you digesting food. Mechanical is movement.

7 0
3 years ago
Read 2 more answers
What is the change in enthalpy for the following reaction? 
Tju [1.3M]

The given chemical reaction is:

2H_{2}O_{2}(aq)---> 2H_{2}O(l)+O_{2}(g)

The standard heats of formation of H_{2}O and H_{2}O_{2} are:

ΔH_{f}^{0}(H_{2}O) = -285.8kJ/mol[/tex]

ΔH_{f}^{0}(H_{2}O_{2}) = -187.6 kJ/mol[/tex]

Calculating the change in heat:

ΔH^{0}_{reaction)=∑ΔH_{f}^{0}(products)-∑ΔH_{f}^{0}(reactants)

                         = [{2 * (-285.8 kJ/mol)} -{2*(-187.6 kJ/mol)}]

                          = -196.4 kJ/mol

Therefore, the change in enthalpy for the given reaction is -196.4 kJ/mol



7 0
4 years ago
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