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Rus_ich [418]
3 years ago
15

The isomers butane and methylpropane differ in their 1) mo;ecular formulas 2) structural formulas 3) total number of atoms per m

olecule 4) total number of bonds per molecule
Chemistry
1 answer:
docker41 [41]3 years ago
7 0

Answer:

Structural formulas

Explanation:

I know cause i just did it

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which of the following true? a_volcanoes and earthquakes often near the plate boundaries. b_volcanoes occur whereve there are ta
pentagon [3]

The answer is A. Volcanoes and earthquakes often occur near place boundaries. Volcanoes are exit points for heat from the Earth's core. And Earthquakes are caused by the tectonic places colliding along another one. For example one tectonic plate would be slowly moving south, and another next to it could collide with the South-going plate while going North, and that is what causes Earthquakes.

4 0
3 years ago
Read 2 more answers
With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?
mina [271]

Answer: 7.07 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} zinc=\frac{21g}{65g/mol}=0.32moles

\text{Moles of} CuCl_2=\frac{7g}{134g/mol}=0.052moles

Zn+CuCl_2\rightarrow Cu+ZnCl_2

According to stoichiometry :

1 mole of CuCl_2 require 1 mole of Zn

Thus 0.052 moles of CuCl_2 will require=\frac{1}{1}\times 0.052=0.052moles  of Zn

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Zn is the excess reagent.

As 1 mole of CuCl_2 give = 1 mole of ZnCl_2

Thus 0.052 moles of CuCl_2 give =\frac{1}{1}\times 0.052=0.052moles  of ZnCl_2

Mass of ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g

Thus 7.07 g of ZnCl_2 will be produced from the given masses of both reactants.

8 0
3 years ago
The town in the video stressed using locally made goods instead of imported goods. How could using locally made goods lower CO2
ladessa [460]

Answer:

It reduces the need to import goods

Explanation:

When you buy locally, the products you buy don't come from far away, so they don't have to cross the country (or the ocean) by boat, plane or trucks to reach the market/store where you're buying, at least not from a long distance away.

The distance a vehicle travels, the less CO2 emissions it produces.

If the good you're buying is made/produced only an hour away, that's not much pollution produced compared as if the good has to come from a distant place spending days on highways to reach you.

6 0
3 years ago
Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to
Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density \rho_{avg}; we have:

\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }

The average atomic weight is:

A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

C_{Fe} = 100 - C_v

From a unit cell volume V_c

V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}

where;

N_A = number of Avogadro constant.

SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

C_{Fe} with 100-C_v

Then:

a^3 = \dfrac   { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) }    {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

Replacing the values; we have:

(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol)   }   }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3)   } }

2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} }  \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }

2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})

\mathbf{C_v = 9.1 \ wt\%}

4 0
3 years ago
An unknown metal displaces cadmium (Cd) from solution but does not displace chromium (Cr). Using the activity series, determine
adell [148]
I found this is the order of activity in Internet

Mn>Zn>Cr>Fe>Cd>Co>Ni>Pb

You can see that Fe is between Cr and Cd

Cr is more active than Fe and Fe is more active than Cd.

The Fe will displace Cd but not Cr.

Answer:Fe
3 0
3 years ago
Read 2 more answers
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