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stepan [7]
3 years ago
6

Pls help and i will mark brainliest

Chemistry
2 answers:
myrzilka [38]3 years ago
6 0

Answer:

No. 3 lithium

Explain that lithium has 3 protons and 3 electrons. There are 2 electrons on the first energy level and 1 electron on the second

No. 4 Hydrogen (H) and helium (He) have a valence shell containing one and two electrons respectively. They make up the first period (row) of the periodic table. Their valence electron/s are in the first energy level (n=1) , as is denoted by 1s1 and 1s2 .

<em>That</em><em>'</em><em>s</em><em> </em><em>all</em><em> </em><em>i</em><em> </em><em>dont</em><em> </em><em>know</em><em> </em><em>if</em><em> </em><em>this</em><em> </em><em>is</em><em> </em><em>right</em><em> </em><em>though</em><em>.</em><em> </em>

Serjik [45]3 years ago
5 0
Sodium
Beryllium
Lithium
Helium
Potassium
Fluorine
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How much energy (heat) is required to convert 248 g of water from 0 oC to 154 oC? Assume that the water begins as a liquid, that
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Answer:

The total heat required is 691,026.36 J

Explanation:

Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L

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On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).

In this case, the total heat required is calculated as:

  • Q  for liquid water.  This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C

Q= c*m*ΔT

Q=4.184\frac{J}{g*C} *248 g* (100 -0 )C

Q=103,763.2 J

  • Q  for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then\frac{248 g}{18 \frac{g}{mol} } =13.78 moles )

Q= m*L

Q=13.78moles*40.79 \frac{kJ}{mol}

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)

  • Q for temperature change from  100.0 ∘ C  to  154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.

Q= c*m*ΔT

Q=1.99\frac{J}{g*C} *248 g* (154 - 100 )C

Q=25,176.96 J

So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J

<u><em>The total heat required is 691,026.36 J</em></u>

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How many grams of nano3 are needed to prepare 100 grams of a 15.0 % by mass nano3 solution? will give brainliest
Greeley [361]

Answer:

Calculate the mass percent of a potassium nitrate solution when 15.0 g KNO3 is dissolved in 250 g

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3. Calculate the weight of table salt needed to make 670 grams of a 4.00% solution.

4. How many grams of solute are in 2,200 grams of a 7.00% solution?

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Mass Percent = Grams of Solute

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Grams of Solute = Grams of Solution X Mass Percent

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Explanation:

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Question 1 of 10
kati45 [8]
I believe the answer is C.
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