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3 years ago

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AdiffusioncoupleofweldedAandBwasgivenadiffusionannealatelevatedtemperatureandthencooledtoroomtemperature.Chemicalanalysisofthesu

ccessivelayercutparalleltotheweldinterfaceshowthatoveradistanceof5000nm,NAchangedfrom0.3to0.35.Assumethatthenumberofatomsperm3ofbothpuremetalsis9×1028.a)determinetheconcentrationgradientdnA/dx,b)ifthediffusioncoefficientatthepointinquestionatannealingtemperaturewas2×10-14m2/s,determinethenumberofAatomspersecondthatwouldpassthroughthiscross-sectionattheannealingtemperature.Neglecttheinterfacemotion.

1 answer:

Murljashka [212]3 years ago

8 0

Answer:

what

Explanation:

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What state of matter having a definite volume but no definite shape

Lubov Fominskaja [6]

The state of matter is liquid.

4 0

3 years ago

Read 2 more answers

If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of

maw [93]

Answer:

False

Explanation:

Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.

<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>

Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.

4 0

3 years ago

Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres

____ [38]

Answer:

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature = $V_1=3.4 L$

Number of moles in the container = $n_1$

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = $V_2=5.11 L$

Number of moles in the container after addition= $n_2=2.28\times 10^{-2} mol$

Applying Avogadro's law:

$\frac{Volume}{Moles}=constant$ (at constant pressure and temperature)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}$

$n_1=1.51\times 10^{-2} mol$

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

8 0

3 years ago

A decay series starts with the synthetic isotope ²³⁹₉₂U. The first four steps are emissions of a β⁻ particle, another β⁻, an a p

VladimirAG [237]

<u>Thorium series</u> could start by this sequence.

<h3>Brief explanation</h3>

To write balanced equations for nuclear decay processes. It's important to remember that the mass number and the atomic numbers must be balanced. And so what that means is that if we look at an elements nuclear symbol, the atomic number is the bottom number and the top number, the superscript, is the mass number, and so when we add them up on both sides, they have to be equal. There are two different ways in which decay can occur.

In this, series one is through beta decay, which means that the following particle is produced. The other is Alpha Decay, which produces this particle. Both are products. So if we start off with uranium to 39 you read it in nuclear notation, which means we have to find the atomic number just 92 and it undergoes beta decay.

So that means that it produces this particle find the second particle we used the atomic number, so 92 equals minus one plus x, where X equals 93 which is Neptune IAM. The mass number of our new isotope is zero plus X equals to 39 where X equals to 39. This product becomes the reactant in my next decay, which is also a beta decay. And to find the unknown element we do the same here.

Except for that it's 93 equals minus one plus x, where X is 94 which is P u plutonium, and the mass number is zero plus X equals to 39 or to 39. The next decay starts with the isotope that we just form to 39 p. U. This time it's an Alpha decay. So we produce this particle to find the unknown. Element 94 equals two plus x, where X equals 92 which takes us back to uranium.

Find the mass number of this isotope 2 39 equals four plus X, where X equals to 35. Finally, for the last decay, you have another Alpha decay starting with uranium to 35 making an alpha particle. The atomic number will be 90 which is T H and the top is 2 31 For the mass number. This begins the natural decay, series of thorium .

Learn more about chemical decay

brainly.com/question/1898040

#SPJ4

7 0

9 months ago

Find percent yield:

saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For $B_5H_9$ :</u>

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

<u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago