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3 years ago

Use the drop-down menus to select the correct name for each of the organic compounds.

Chemistry

2 answers:

irakobra [83]3 years ago

6 0

Answer:

edge 2020

Explanation:

Molodets [167]3 years ago

5 0

Answer:

Here's what I get

Explanation:

CH₃CH₂CH₂CH₂CH₂CH₃ — hexane

CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted

CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane

CH₃CH₂CHCICH₂CH₃ — 3-chloropentane

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An important reaction that takes place in a blast furnace during the production of iron is the formation of iron metal and CO2 f

ladessa [460]

Answer: Mass of $Fe_2O_3$ required to form 930 kg of iron is 1328 kg

Explanation:

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

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

For iron:

Given mass of iron = 930 kg = 930000 g (1kg=1000g)

Molar mass of iron = 56 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron}=\frac{930000g}{56g/mol}=16607mol$

The chemical equation for the production of iron follows:

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

By Stoichiometry of the reaction:

2 moles of iron are produced by = 1 mole of $Fe_2O_3$

So, 16607 moles of iron will be produced by = $\frac{1}{2}\times 16607=8303moles$ of $Fe_2O_3$

Now, calculating the mass of $Fe_2O_3$ from equation 1, we get:

Mass of $Fe_2O_3$ = $moles\times {\text {molar mass}}=8303\times 160=1328480g=1328kg$

Thus mass of $Fe_2O_3$ required to form 930 kg of iron is 1328 kg

6 0

3 years ago

Complete the table below by deciding whether a precipitate forms when aqueous solutions A and B are mixed. If a precipitate will

stiks02 [169]

The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.

The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.

Now, we will consider each reaction individually to decode whether or not a precipitate is possible.

In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.

Learn more: brainly.com/question/11969651

6 0

3 years ago

Help plzzzzzzz ASAP!

Agata [3.3K]

I am going to say C. it has to do with the angles

8 0

3 years ago

Read 2 more answers

1Which is a dopant for a p-type semiconductor?

aalyn [17]

For a p type of semiconductor we need a dopant which is from 13th group in periodic table

Al , B, Ga, In Tl

So the correct element will be In : Indium

The other elements belongs to 15th group and hence will give n type semiconductor

5 0

3 years ago

Read 2 more answers

A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure

GenaCL600 [577]

Answer:

$P_A=4.20atm\\\\P_B=17.1atm$

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

$PV=nRT$

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

$P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm$

Best regards!

8 0

3 years ago