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Lena [83]
3 years ago
5

The temperature at which the vapor pressure of a liquid is equal to the external pressure is called the _____ point.

Chemistry
1 answer:
jeyben [28]3 years ago
4 0
It is B. Boiling point
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When cleaning up a local beach, students found many different particles that were in the water that affected the shoreline, like
ddd [48]

Your answer is C. Both gasoline and litter would need to be physically separated from the water, because neither bonds with the water.


8 0
3 years ago
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A sample of 0.370 mol of a metal oxide (m2o3) weighs 55.45 g. how many grams of o are in the sample?
Wewaii [24]
0.370 mol metal oxide = 55.45 g 

<span>1 mol = 55.45/0.370 = 149.86 g </span>

<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>

<span>there is 48/149.86 * 100% O in the sample </span>

<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
6 0
3 years ago
In the nucleus with the protons together their sum is the
galina1969 [7]

Answer:

the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom

Explanation:

6 0
3 years ago
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In the following reaction, how many grams of ammonia (NH3) will react with 27.8 grams of nitric oxide (NO)? 4NH3 + 6NO → 5N2 + 6
Rzqust [24]
<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O

mol of NO  =  </span>\frac{mass of NO}{Molar Mass of NO}
 
                  =  \frac{27.8 g}{30.01 g / mol}
                   
                  =  0.93 mol

Based on the balance equation mole ratio of NH₃  :  NO  is   4 : 6
                                                                                            =  2 : 3

If mol  of NO  =  0.93 mol

then mol of NH₃ = \frac{0.93 mol  *  2}{3}
                           
                          =  0.62 mol

Mass of ammonia =  mol  ×  molar mass
          
                             =  0.62 mol   ×  17.03 g/mol
 
                             =  10.54 g

Therefore B is the best answer





3 0
3 years ago
Read 2 more answers
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
3 years ago
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