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3 years ago

10

Which statement correctly compares an atom of an alkali metal with an atom of the alkaline earth metal next to it on the periodi

c table?

1 answer:

Mumz [18]3 years ago

6 0

<span>The alkali metal atom forms a +1 ion, while the alkaline earth metal atom forms a +2 ion. hope this helps</span>

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What atomic or hybrid orbitals make up the sigma bond between br and f in bromine trifluoride, brf3?

mina [271]

<span>Draw the Lewis structure. Bromine has 3 bonds and two lone pairs for a trigonal bipyramidal electron geometry and an sp^3d hybridization. Fluorine is peripheral it does not require hybridization, but we often consider it to be hybridized too - it has 1 bond and 3 lone pairs for sp^3 hybridization. So the sigma bonds come from an overlap of an sp^3d orbital on Br with an sp^3 orbital on F. If you don't consider the F to be hybridized the overlap would have to be to a p orbital on the F</span>

8 0

4 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

As water changes state, the water either absorbs or releases energy. Which of these

USPshnik [31]

Answer:

a, ocean water evaporating

6 0

4 years ago

For the reaction of hydrogen with iodine

fenix001 [56]

Answer:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Explanation:

Hello!

In this case, considering the given chemical reaction:

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

Thus, by applying the law of rate proportions, we can write:

$\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}$

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Best regards!

8 0

3 years ago

Calculate the mole fraction of the total ions in an aqueous solution prepared by dissolving 0.400 moles of MgBr2 in 850.0 g of w

diamong [38]

<span>moles water = 850.0 g / 18.02 g/mol=47.2
moles Mg2+ = 0.400
moles Cl- = 2 x 0.400 = 0.800
moles ions = 0.400 + 0.800= 1.2
mole fraction ions = 1.2 / 1.2 + 47.2 =0.0248</span>

7 0

3 years ago

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