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Brilliant_brown [7]
4 years ago
6

In which way does a na1+ ion differ from a neutral na atom?

Chemistry
1 answer:
DochEvi [55]4 years ago
8 0
Na 1+ misses 1 electron
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Great Sand Dunes National Park in Colorado is famous for its giant sand dunes. Sand dunes are landforms that are found in desert
Anit [1.1K]
Either b or c :), i’m not quite sure
5 0
3 years ago
Which of these formulas is NOT consistent with the guidelines for creating ionic compounds? Mark all that apply.
iVinArrow [24]

Answer:

d

Explanation:

because boron and fluorine are both nonmetals and don't fit the guidlelines for creating ionic compounds

8 0
3 years ago
How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

4 0
3 years ago
At 580 k, 6. 42% of the molecules are in the chair form. calculate the value of the equilibrium constant for the reaction as wri
Ann [662]
Equilibrium chemical reaction between chair and boat forms is:
Kc = \frac{[Boat form]}{[Chair form]}
Suppose the total number of molecules is x 
Number of molecules of chair form is 6.42/100 x
Number of molecules of boat form is 93.58/100 x
Kc = \frac{(0.9358x)}{(0.0642x)} = 14.6
5 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
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