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Cloud [144]
3 years ago
6

if the temperature of 20.0 L of gas were increased to twice the initial value, while the pressure is held constant what would be

the volume of the gas at the new temperature
Chemistry
1 answer:
Kay [80]3 years ago
7 0

Answer:

40 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 20 L

Initial temperature (T₁) = T

Final temperature (T₂) = 2T

Pressure (P) = constant

Final volume (V₂) =?

The new (i.e final) volume of the gas can be obtained as follow:

V₁/T₁ = V₂/T₂

20/T = V₂/2T

Cross multiply

T × V₂ = 20 × 2T

T × V₂ = 40T

Divide both side by T

V₂ = 40T /T

V₂ = 40 L

Therefore, the new volume of the gas is 40 L

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In the equation:
Ksenya-84 [330]

Answer:

B) 16 g

Explanation:

  • 2H₂ + O₂ → 2H₂O

First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 4 mol O₂ * \frac{2molH_2}{1molO_2} = 8 mol H₂

Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:

  • 8 mol H₂ * 2 g/mol = 16 g

Thus, the correct answer is option B).

6 0
3 years ago
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

8 0
2 years ago
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
The reactant concentration in a first-order reaction was 7.60 x 10-2 M after 35.0 s and 5.50 x 10-3 M after 85.0 s. What is the
8090 [49]

Answer:

k = -0.0525 s⁻¹

Explanation:

The equaiton for a first order reaction is stated below:

ln[A]=−kt+ln[A]₀.

[A] = 5.50 x 10⁻³ M

[A]₀ = 7.60 x 10⁻² M

t = 85.0 - 35.0 = 50.0 s

The rate constant is represented by k and can be calculated substituting the values given above:

k = (ln[A]₀ - ln[A])/t

k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s

k = -0.0525 s⁻¹

4 0
2 years ago
Read 2 more answers
What is the mass percent of a sugar solution when 489 grams of sugar is combined with 877 grams of water?​
natka813 [3]
The sugar solution was made by mixing 12 grams of water
4 0
2 years ago
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