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2 years ago

I'm not sure what the answer is

Chemistry

1 answer:

Talja [164]2 years ago

4 0

Answer:

I think is A

Explanation: i'm not sure if im right.

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I need this ASAP thank you

Rainbow [258]

Answer:

16,,24Mg 17,,a24.1 18a mass number of the most abundant isotope

Explanation:

atomic number of Mg is 12 ,therefore its mass number should be the value that is very close to 24.

24.1 is the value of thee most abundant isotope.

8 0

2 years ago

Consider the following reaction.Cr2O3(s) + 3CCl4(l) 2CrCl3(s) + 3COCl2(g). When the green solid is mixed with the colorless liq

Elena L [17]

Answer:

The answer to your question is: letter B

Explanation:

Reaction

Cr2O3(s) + 3CCl4(l) ⇒ 2CrCl3(s) + 3COCl2(g)

From the information given and the reaction, we can conclude that:

Green solid = Cr2O3 (s) "s" means solid

Colorless liquid = CCl4 (l) "l" means liquid and is the other reactant

Purple solid = CrCl3(s) CrCl3 is purple and "s" solid

Then, as a green specks remains it means that the excess reactant is Cr2O3, so, CCl4 is the limiting reactant.

6 0

3 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

Why can you never see the far side of the mood from earth?

Yuliya22 [10]

Answer: -

C) because the moon rotates once on its axis in the same time that it takes to revolve around the earth

Explanation: -

The moon rotates about it's axis one full time at about the same time as it takes to rotate around the earth.

This leads to only one side of the moon always appearing to Earth, called the near side.

One side of the moon is always not seen from the earth. It is called the far side of the moon.

Thus we can never see the far side of the moon because the moon rotates once on its axis in the same time that it takes to revolve around the earth

4 0

3 years ago

Why should scietist as questions

larisa86 [58]

Answer:

So they have the right answers

Explanation:

I know

7 0

3 years ago