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9 months ago

1 question content area what is the total number of valence electrons in the lewis structure of sebr2o?

Chemistry

1 answer:

joja [24]9 months ago

6 0

There are 24 valance electrons in the compound.

<h3>What are valence electrons?</h3>

We know that the valence electrons are those that are found on the valence shell of an atom. There are four atoms that can be seen in the compound that is under consideration. The atoms are selenium, bromine (2 atoms) and oxygen.

Given that there are seven valence electrons in bromine and six valence electrons in oxygen and selenium. We have;

2(7 electrons of bromine) + 6 electrons of oxygen + 4 electrons of Se

= 24 electrons

Learn more about valence electrons:brainly.com/question/8906371

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7. A student has a 25 g sample of gas in a container at STP and adds another 7

BigorU [14]

Answer:

New temperature t2 = [1.28T− 273.15]° C

Explanation:

Given:

Volume v1 = 25 gram

New volume v2 = 25 + 7 = 32 gram

Constant pressure = p

Temperature t1 = T

Find:

New temperature t2

Computation:

Pv1/t1 = Pv2/t2

25 / T = 32 / t2

t2 = 1.28T

New temperature t2 = [1.28T− 273.15]° C

4 0

2 years ago

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

4 0

2 years ago

How many moles of ammonla are produced by 2.8 mol of hydrogen.<br><br>ANH

yan [13]

Answer:

3 i think

Explanation:

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2 years ago

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3 years ago

How much force is needed to accelerate a 66 kg skier at 2 m/sec2

alexgriva [62]

Answer:

132N

Explanation:

F=ma

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F=132N

7 0

3 years ago