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elena-s [515]
3 years ago
12

QUESTION 5

Chemistry
2 answers:
Black_prince [1.1K]3 years ago
5 0
I think A is the correct answer
taurus [48]3 years ago
3 0

Answer:

Beta Rays

Beta Rays are not Electromagnetic Waves

Beta rays also known as beta radiation is obtained through the emission of an electron. Beta rays are not electromagnetic waves because they are charged particles and are capable of getting deflected by the magnetic field.

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As an object falls freely near the earths surface , the loss in gravitational potential energy of the object is equal to its wha
Nitella [24]
Increase in kinetic energy as well as energy loss to the surroundings in the form of heat ( negligible)
4 0
3 years ago
Read 2 more answers
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
3. Five grams of orange-drink mix are
navik [9.2K]

Answer:

I think it is C 105

Explanation:

tell me i am worong

5 0
3 years ago
Dissolve 30 g of sodium sulphate into 300 mL of water
Aneli [31]

Answer:

number of moles = 0.21120811

Explanation:

To find the number of moles, given the mass of the solute, we use the formula:

\mathrm{n =   \dfrac{ m  }{ M  } }

\mathrm{n = number\:of\:moles\:(mol)}

\mathrm{m = mass\:of\:solute\:(g)}

\mathrm{M = molar\:mass\:of\:solute\:(  \dfrac{ g  }{ mol  }   )}

Label the variables with the numbers in the problem:

\mathrm{n =\:?}

\mathrm{m =30\:g }

\mathrm{M =\:?\:Calculate\:the\:molar\:mass }

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

M({ \left Na \right }_{ 2  }   { \left So \right }_{ 4  })   =  m \left( Na  \right)  +m \left( S  \right)  +m \left( O  \right)

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

\mathrm{M =molar\:mass }

\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}

\mathrm{Na =sodium=22.99\:g }

\mathrm{S =sulfur=32.06\:g }

\mathrm{O =oxygen=16.00\:g }

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

M({ \left Na \right }_{ 2  }   { \left So \right }_{ 4  })   =  m \left( Na  \right)  +m \left( S  \right)  +m \left( O  \right)

\mathrm{Substitute\:the\:values\:into\:the\:formula}

M  =  2 \left( 22.99  \right)  +1 \left( 32.06  \right)  +4 \left( 16.00  \right)

\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}

\mathrm{M =  45.98+32.06+4\:(16)}

\mathrm{Multiply\:4\:by\:16\:to\:get\:64}

\mathrm{M =  45.98+32.06+64}

\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}

\mathrm{M =  78.04+64}

\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}

\mathrm{M =  142.04}

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

\mathrm{n =   \dfrac{ m  }{ M  } }

\mathrm{n =\:?}

\mathrm{m =30\:g }

\mathrm{M = 142.04\:g/mol}

\mathrm{Substitute\:the\:values\:into\:the\:formula}

\mathrm{n =   \dfrac{ 30  }{ 142.04  }}

\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}

\mathrm{n =  0.21120811}

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

3 0
2 years ago
Vinegar, fruit juice and cola are examples of
andrew-mc [135]

Answer:

strong bases

Explanation:

7 0
3 years ago
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