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3 years ago

Help PlS AFAP and thank you so much

Chemistry

2 answers:

Triss [41]3 years ago

8 0

Answer:

Explanation:

idk how to explain, B is the definition of conduction

Umnica [9.8K]3 years ago

8 0

the answer is B transfer thru direct contact

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For which characteristic would a human MOST LIKELY selectively breed a house cat?

madreJ [45]

B. Coat color I’m pretty sure

8 0

3 years ago

Read 2 more answers

Please Help, will give 30 points! The following data was collected when a reaction was performed experimentally in the laborator

Sedaia [141]

Answer:

9 moles of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaCl —> 3NaNO3 + AlCl3

Next, we shall determine the number of mole of Al(NO3)3 and NaCl that reacted and the number of mole of NaNO3 produced from the balanced equation.

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl to produce 3 moles of NaNO3.

Next, we shall determine the limiting reactant.

The limiting reactant can be obtained as follow:

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl.

Therefore, 4 moles of Al(NO3)3 will react with = (4 x 3)/1 = 12 moles of NaCl.

From the calculations made above, we can see that it will take a higher amount i.e 12 moles than what was given i.e 9 moles of NaCl to react completely with 4 moles of Al(NO3)3.

Therefore, NaCl is the limiting reactant and Al(NO3)3 is the excess reactant.

Finally, we shall determine the maximum amount of NaNO3 produced from the reaction.

In this case, the limiting reactant will be used as it will produce the maximum amount of NaNO3 since all of it is consumed by the reaction.

The limit reactant is NaCl and the maximum amount of NaNO3 produced can be obtained as follow:

From the balanced equation above,

3 moles of NaCl reacted to produce 3 moles of NaNO3.

Therefore, 9 moles of NaCl will also react to produce 9 moles of NaNO3.

From the calculations made above, the maximum amount of NaNO3 produced is 9 moles

6 0

3 years ago

List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)

Leokris [45]

Answer:

20, 44, 62

Explanation:

To find the number of atoms of each element, we multiply coefficient and subscript

For example $5 Ca_1 Cl_2$ contains

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example $3Ca_3 (P_1 O_4 )_2$

we multiply coefficient × subscript × number outside the bracket to find the number of atoms

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

3 × 3 = 9 , Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present.

So let us find the number of atoms of each element on the left side of the equation

$2C_{10} H_{22} (l)+31O_2 (g)\Rightarrow 20CO_2 (g)+22H_2 O(g)$

Number of C atoms = 2 × 10 = 20

Number of H atoms = 2 × 22 = 44

Number of O atoms = 31 × 2 = 62

20, 44, 62 are the Answers.

3 0

3 years ago

What is the correct name for MgF2?

ANTONII [103]

MgF₂ ( Magnesium fluoride )

hope this helps!

3 0

3 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

3 years ago