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3 years ago

Number of electrons in the outer ring of bromine

1 answer:

8 0

Simple,

looking at the periodic table..

1s2,2s2,2p6,3s2,3p6,4s2,3d10,4s5

Thus, there are 5 valence electrons in the outer ring of Bromine.

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Which produces oxygen-18 when it undergoes beta decay? A. oxygen-17 B. oxygen-18 C. fluorine-17 D. fluorine-18

inna [77]

Answer:

Option D. fluorine-18

Explanation:

The attached photo gives the explanation.

In the attached photo, A zX represent the atom that will undergo beta decay to produce oxygen–18.

After the calculation, A zX was found to be fluorine–18.

3 0

3 years ago

Which statement describes the BEST example of animals avoiding competition?

Svetlanka [38]

Answer:

Explanation:

If the deer leave the forest then they are avoiding compotation

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2 years ago

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Why do plants need sap

Vinil7 [7]

Plants don't need sap. Plants produce sap and attract bugs.

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3 years ago

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write the steps you would use for making tea.use the words solution,solvent,solute,dissolve,soluble,insoluble,filtrate and resid

poizon [28]

1. Take 100ml of water as solvent and boil it few minutes.
2. Now add one tea spoon sugar, one tea spoon tea leaves and 50ml of milk. Here sugar, tea leaves and milk are solute.
3. Now boil it again for few minutes so that sugar will dissolves in solution as sugar is soluble in water
4. Now filter the solution. Collect the filtrate in cup. The insoluble tea leaves will be left behind as residue.

4 0

3 years ago

If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?

Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 412.072 \ g \ Cl_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

412.072 g Cl₂ ≈ 412 g Cl₂

5 0

3 years ago