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Sloan [31]
3 years ago
14

Part 1: What is a black dwarf? Part 2: What is the fate of Sol?

Chemistry
1 answer:
irinina [24]3 years ago
5 0
Part 1 : a black dwarf is a theoretical stellar remnant part: 2 idk
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Which statements are true of geothermal energy? Check all that apply.
salantis [7]

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Explanation:

6 0
3 years ago
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What is the order in which electrons start filling the orbitals?
earnstyle [38]
<span>Take a look at this periodic table. You start in the left upper corner (1s) then you go to the right untill you can't go further, then you go 1 row down and start at the left again. So the order will be 1s,2s,2p,3s,3p,4s,3d,4p... etc</span>
5 0
3 years ago
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Which is the term for an ion that has more protons than electrons
murzikaleks [220]

Answer:

The correct answer is cation. See the explanation below, please.

Explanation:

An atom with a net charge, either positive or negative, is called an ion. In the event that an atom loses an electron (or more), that is, it will have more protons than electrons, and its net charge will be positive, it will be called cation. In the opposite direction, if an atom gains electrons, it will have a negative net charge, called anion.

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
NaHCO 3 and NaHSO 4 are acid salts. however aqueous solution of NaHCO 3 is basic while aqueous solution of NaHSO 4 is acidic. gi
posledela

Answer:c

Explanation:

3 0
3 years ago
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