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2 years ago

14

30 POINTS!! Will mark brainliest if correct answer and all work is shown! What mass in grams of tin would be required to complet

ely react with 1.30 L of 0.750 M HBr in the following chemical reaction? Sn(s) + 4 HBr(aq) → SnBr₄ (aq) + 2 H₂ (g)

1 answer:

olya-2409 [2.1K]2 years ago

3 0

Answer:

mSn=0.24375*59=14.38125

Explanation:

nHBr=1.3*0.75=0.975

nSn=1/4nHBR=0.975/4=0.24375

mSn=0.24375*59=14.38125

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the red line observed in the line spectrum for hydrogen has 3.03x10^-19J, what is the wavelength, in nm, of this?

djverab [1.8K]

Answer:

λ = 6.5604 x 1016 nm

Explanation:

Given Data:

The energy of the red line in Hydrogen Spectra = 3.03 x 10-19

Formula to calculate Wave length

E= hv

Where E is Energy

h is Planks Constant = 6.626 x 10–34 J s

v is frequency

In turn

v= c/ λ

where c is speed of light = 3.00 x 108 m s–1

λ is wavelength = to find

Solution:

Formula to be Used:

E= hv………………………… (1)

Putting the value v in equation 1

E= h c/ λ…………………… (2)

Put the value in equation 2

3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)

By rearranging equation 3

λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J

λ = 6.5604 x 107 m

The answer is in “m”

So we have to convert it into nm

So for this to convert “m” to “nm” multiply the answer with 109

λ = 6.5604 x 107 x 109

λ = 6.5604 x 1016 nm

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2 years ago

Intermolecular forces dipole differences london dispersion

loris [4]

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2 years ago

Write the net ionic equation for the precipitation of cobalt(ii) carbonate from aqueous solution

andrey2020 [161]

Missing question: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of ammonium carbonate and cobalt(II) bromide are combined.<span>Balanced chemical reaction:
(NH</span>₄)₂CO₃(aq) + CoBr₂(aq) → CoCO₃(s) + 2NH₄Br(aq).
Net ionic reaction:
2NH₄⁺(aq) + CO²⁻(aq) + Co²⁺(aq) + 2Br⁻(aq) → CoCO₃ + 2NH₄(aq)+ 2Br(aq).
or CO²⁻(aq) + Co²⁺(aq) → CoCO₃(s).

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

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2 years ago

In a chemical change, the substance that you end up with is the

elena55 [62]

The reactant is what you begin with.
The product is what you end up with (so the answer is B)

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2 years ago

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