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3 years ago

What is the oxidation number of Br in BrCl

1 answer:

6 0

By definition, oxidation number is the charge left on the given atom when all the bonding pairs (of electrons) are broken, so the oxidation number of Br will be +1

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The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

What is the average kinetic energy and rms speed of N₂ molecules at STP? Compare these values with those of H₂ molecules at STP.

Otrada [13]

The average kinetic energy and rms speed of N₂ molecules at STP is $5.65686 \times 10^{-21}$ and $493 \mathrm{~m} / \mathrm{s}$

Given,

$\begin{aligned}&\mathrm{P}=1.013 \times 10^{5} \mathrm{~Pa} \\&\mathrm{~T}=273.15 \mathrm{~K} \\&\rho=1.25 \mathrm{~kg} / \mathrm{m}^{3}\end{aligned}$

The average kinetic energy of a molecule is given by, $K . E .=\frac{3}{2} k T$ where k is the Boltzmann constant and Tis the absolute temperature of the gas.

$K . E .=\frac{3}{2} \times $1.380649 \times 10^{-23}$ \times 273.15 K$

$K.E=$5.65686 \times10^{-21}$$

The rms speed of $\mathrm{N}_{2}$ molecules is given by

$\begin{aligned}&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \times 1.013 \times 10^{5}}{1.25}}=493 \mathrm{~m} / \mathrm{s}\end{aligned}$

The average kinetic energy of a gas's particles is inversely related to its temperature. As the gas warms, the particles must travel more quickly since their mass is constant.

The average kinetic energy (K) is equal to one half of the mass (m) of each gas molecule times the RMS speed (vrms) squared.

Learn more about average kinetic energy brainly.com/question/1599923

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6 0

2 years ago

The following equation is balanced correctly. 4C2H6 + 7O2 --> 6H2O + 4CO2 True False

Ostrovityanka [42]

Answer: False

Explanation:

4C2H6 + 7O2 --> 6H2O + 4CO2

8 Carbons on the reactant side, but 4 Carbons on the Product

24 Hydrogens on Reactant, 12 H on product

But Oxygen is balanced, 14 on each side

4 0

3 years ago

Read 2 more answers

HELP!!!!!!!!!! PLEASE!!! HELP!!

enyata [817]

Answer:

Ooh sweetie goodluck

Explanation:

May the gods ever be in your favor.

8 0

3 years ago

If it take 20 second for 200cm3 of H25 to diffuse through a porovs pot,hours,long will it take 400cm3 of NH3 of diffuse under th

trapecia [35]

Answer:

40 seconds chief just double it

Explanation:

5 0

3 years ago