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AURORKA [14]
3 years ago
10

20 points be right this is my final!!!

Chemistry
2 answers:
nevsk [136]3 years ago
6 0
I think it’s c hope it helps.
Andreyy893 years ago
4 0
What else will happen.
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3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.
tiny-mole [99]

Answer:

The Avogadro's  number is N_A     =  6.02289 *10^{23}

Explanation:

From the question we are told that

   The edge length is  L   = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100}  =  5.02 * 10^{-10}

    The density of the metal is \rho =  5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3}  * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3

     The molar mass of  Ba is  Z  =  137.3 \ g/mol = \frac{137.3}{1000} =  0.1373 \  kg / mol

     

Generally the volume of a unit cell is  

       V =  L^3

substituting value

        V =  [5.02 *10^{-10}]^3

         V = 1.265*10^{-28}\ m^3  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

The volume of barium atom is  

        V_a  =  \frac{V}{2}  * 0.68

substituting value

        V_a  =  \frac{ 1.265*10^{-28}}{2}  * 0.68

        V_a  = 4.301 *10^{-29} \ m^3

The Molar mass of barium is mathematically represented as

      Z  =  N_A V_a *  \rho

Where N_A is the Avogadro's number

 So  

      N_A     =   \frac{ Z}{ V_a *  \rho}

substituting value

     N_A     =   \frac{ 0.1373}{ 4.301*10^{-29} *  5.3*10^{3}}

     N_A     =  6.02289 *10^{23}

4 0
3 years ago
A patient receives 3.3 L of glucose solution intravenously (IV). If 100. mL of the solution
artcher [175]

Answer:

660kcal

Explanation:

The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.  

We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:  

3.3L * (1000mL / L) * (5g/100mL)= 165 g.

If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.

6 0
3 years ago
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

 

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

 

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

 

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

6 0
3 years ago
Image "B" shows more biodiversity than image "A". Why?
Ilya [14]

Answer:

The answer is A

Explanation:

Image A has the same species while image B has more species than A.

8 0
3 years ago
Which of the following has the correct name for either the acid or base
Vlad1618 [11]

Answer:

Explanation:

harmless

...............

6 0
3 years ago
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