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Whitepunk [10]
3 years ago
9

Can someone help me?

Chemistry
2 answers:
elena-s [515]3 years ago
4 0
It’s will decrease I have to make it 20 characters but it’s decrease
Slav-nsk [51]3 years ago
3 0
The answer is it will decrease
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I’LL MAKE YOU BRAINLIEST+ FREE POINTS
vladimir2022 [97]

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

Ci*Vi=Cf*Vf

where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.

In this case, being:

  • Ci= 7 M
  • Vi= 0.60 L
  • Cf= 2.5 M
  • Vf=?

Replacing:

7 M*0.60 L= 2.5 M* Vf

Solving:

Vf=\frac{7 M*0.60 L}{2.5 M}

Vf= 1.68 L

<u><em>The solution's new volume is 1.68 L</em></u>

6 0
3 years ago
Mantle material rises in convection currents because heated materials becomes more dense
sergey [27]
Heated mater rises and cold mater sinks
3 0
3 years ago
The monomer of poly(vinyl chloride) has the formula C2H3Cl. If there are 1,565 repeat units in a single chain of the polymer, wh
monitta

Answer:

\large \boxed{9.780 \times 10^{4}\text{ u}}

Explanation:

The molecular mass of a monomer unit is:

C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u

For 1565 units,

\text{Molecular mass} = \text{1565 units} \times \dfrac{\text{62.494 u}}{\text{1 unit }} = \mathbf{9.780 \times 10^{4}}\textbf{ u}\\\\\text{The molecular mass of the chain is $\large \boxed{\mathbf{9.780 \times 10^{4}}\textbf{ u}}$}

8 0
3 years ago
How many atoms are in 0.230grams Pb
Brilliant_brown [7]
Its about 11.5hg because if you divide it with the atom it would result to 11.5hg
5 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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