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brilliants [131]
2 years ago
8

Assessment timer and count

Chemistry
2 answers:
beks73 [17]2 years ago
4 0
Water
sunlight
and carbon dioxide
Maslowich2 years ago
3 0
A plant needs Sunlight, water, and carbon dioxide to produce ATP from photosynthesis.
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What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques?
amid [387]

Answer:

The principal elements are Iron, Nickel, Chromium

It has 0.37% to 0.43% Carbon

It is hardenable.

It can be hardened by cold working, annealing, quenching.

Explanation:

A)

The chemical composition of AISI 4340 Steel are as follows:

Iron  ---  95% to 96%

Nickel  ---  1.6% to 2.0%

Chromium  ---  0.7% to 0.9%

Manganese  ---  0.6% to 0.8%

Carbon  ---  0.37% to 0.43%

Molybdenum  ---  0.2% to 0.3%

Silicon  ---  0.15% to 0.30%

Sulfur  ---   0.040%

Phosphorous  --- 0.0350%

So, the principal alloy elements from this composition are <u>Iron, Nickel and Chromium</u>

B)

The carbon content in this alloy is <u>0.37% to 0.43%</u>

C)

<u>Yes, it can be hardened</u>.

D)

<u>It can be Hardened by Cold Working, Annealing or Quenching</u>.

3 0
2 years ago
For the following problem convert both the reactants to moles and balance chemical equationsThe reaction of 167 g Fe2O3 with 85.
saul85 [17]

Let's start by balancing the reaction:

Fe_2O_3+CO\longrightarrow Fe+CO_2

As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

Fe_2O_3+3CO\longrightarrow Fe+3CO_2

That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

Fe_2O_3+3CO\longrightarrow2Fe+3CO_2

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

M_{Fe_2O_3}=2\cdot M_{Fe}+3\cdot M_O=(2\cdot55.845+3\cdot15.9994)g/mol=159.6882g/molM_{CO}=1\cdot M_C+1\cdot M_O=(1\cdot12.0107+1\cdot15.9994)g/mol=28.0101g/mol

Now, we can convert their masses to number of moles:

\begin{gathered} M_{Fe_{2}O_{3}}=\frac{m_{Fe_2O_3}}{n_{Fe_{2}O_{3}}} \\ n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_{2}O_{3}}}=\frac{167g}{159.6882g/mol}=1.045787\ldots mol \end{gathered}\begin{gathered} M_{CO}=\frac{m_{CO}}{n_{CO}} \\ n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{85.8g}{28.0101g/mol}=3.063180\ldots mol \end{gathered}

Now, to determine the limiting reactant, we need to divide both the number of mole by their coefficients on the balanced reaction, so we can see how many we need per reaction of each:

\begin{gathered} Fe_2O_3\colon\frac{n_{Fe_2O_3}}{1}=\frac{1.045787\ldots mol}{1}=1.045787\ldots mol \\ CO\colon\frac{n_{CO}}{3}=\frac{3.063180\ldots mol}{3}=1.021060\ldots mol \end{gathered}

Now, the limiting reactant is the one we have less number of moles per reaction. We can see that we have less CO than Fe₂O₃, so the limiting reactant is CO.

4 0
1 year ago
Calculate the amount of mole of iron produced from the reaction of 15.9 grams of iron oxide.
Usimov [2.4K]

Answer:

0.19875

Explanation:

nFe2O3=0.099375

nFe=2nFe2O3=0.19875

3 0
2 years ago
What is the thinnest type of crust?
Trava [24]
The correct answer is oceanic crust, 80 km, Hope this helps let me know.
8 0
3 years ago
Will a horse galloping in a field produce and average velocity of zero
motikmotik

Answer:

No

Explanation:

Depending on the mass of the horse and the speed, velocity will change.

5 0
3 years ago
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