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marissa [1.9K]
2 years ago
5

An element that has 2 core electrons and 3 valence electrons

Chemistry
1 answer:
gavmur [86]2 years ago
6 0

Answer:

Boron

Explanation:

You can find this by looking at the number of protons in Boron, 5.

Then calculate how many electrons you are given, in this case the 2 core plus the 3 valence equal 5 total electrons

Neutral elements have the same number of protons and electrons, so your answer would be the element with 5 electrons, Boron.

You can also know this by using electron configuration. Since you kow there are 5 electrons then you can use EC to find out where your element is. In this case it is: 1s2 2s2 2p1

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I need help with this, please :00000
mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of  N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if          1 mole of N₂ produce 2 moles of NH₃

then    3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

Learn more about:

limiting reactant

brainly.com/question/13979150

#learnwithBrainly

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students used a balance and a graduated cylinder to collect the data shown in table 7. calculate the density of the sample. if t
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Answer:

              Percentage error  =  1.88 %

Solution:

Data Given:

                 Mass of Sample  =  20.46 g

                 Volume of Sample  =  43.0 mL - 40.0 mL  =  3.0 mL

Formula Used:

                 Density  =  Mass / Volume

Putting values,

                 Density  =  20.46 g /  3.0 mL

                 Density  =  6.82 g.mL⁻¹

Percentage Error:

                 Experimental Value  =  6.82 g.mL⁻¹

                 Accepted Value  =  6.95 g.mL⁻¹

                 = 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100  =  98.12 %

                 Percentage Error  =  100 % - 98.12 %

                Percentage error  =  1.88 %

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