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3 years ago

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How old are the lunar highlands? 1. 1.8 to 2.6 billion years old 2. 3.1 to 3.8 billion years old 3. 4.0 to 4.3 billion years old

4. less than 1 billion years old?

2 answers:

Anastaziya [24]3 years ago

4 0

They are about 4.5 billion years old. Hope this helps.

irinina [24]3 years ago

3 0

Answer:

they are 4.3 billion years old

Explanation:

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7. The youngest part of the ocean floor is found ____.

aev [14]

Answer:

The answer is Near The Ocean Ridges.

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2 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

2 years ago

What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)

Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

$Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\ ml\\\\V = 33.33 \ ml$

Therefore, volume required is 33.33 ml .

5 0

3 years ago

What would be the effect on the observed melting point of sample were poorly packed?

Neko [114]

If a sample is packed poorly, the sample will not heat evenly and will take longer to melt.

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3 years ago

at standard pressure, what is the temperature at which a saturated solution of NH4Cl has a concentration of 60g NH4CL/100 g H2O

Nataliya [291]

Answer: Temperature = T, unknown

Saturated Solution, NH4Cl concentration = 60g/100g H2O = 0.6g NH4Cl/g H2O

Assume density of H2O = 1 g/ml

m = 0.6g NH4Cl/g H2O / 1 g/ml

m = 0.6g NH4Cl/ml

See the table of saturated solutions and identify the temperature at which the concentration of NH4Cl is 60g/100g H2O.

Explanation: The line on the graph on reference table G indicates a saturated solution of NH4CL as a concentration of 60. g NH4 Cl/100. g H2O

5 0

1 year ago