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2 years ago

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Magnesium ions for ionic bonds with fluoride ions in a 1:2 ratio. Explain how electrons are transferred between atoms and how th

e ionic bonds form in this compound.

1 answer:

kiruha [24]2 years ago

4 0

I hope this helps......

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You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m

jeka94

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

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2 years ago

PLEASE ANSWER ASAP

Sergio [31]

Answer:

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2 years ago

Read 2 more answers

2. A shopper in a supermarket takes a box of sugar from a shelf that is 1.5 m high because he is

muminat

Explanation:

kinetic energy?? idek hope I helped in anyway possible

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3 years ago

Which postulate of Dalton's atomic theory is the result of the law of conservation of mass

creativ13 [48]

The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is: Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

8 0

3 years ago

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

4 0

2 years ago