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dem82 [27]
3 years ago
14

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

late the volume of the 1.475 M stock NaOH solution needed to prepare 250.0 mL of 0.1374 M dilute NaOH solution.
Chemistry
1 answer:
Svet_ta [14]3 years ago
8 0

Answer:

V_1=23.3~mL

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

C_1*V_1=C_2*V_2

Now we can identify the variables:

C_1=~1.475_M

V_1=~?

C_2=~0.1374_M

V_2=~250.0~mL

If we plug all the values into the equation:

1.475_M*V_1=0.1374_M*250.0~mL

And we solve for V_1:

V_1=\frac{0.1374_M*250.0~mL}{1.475_M}

V_1=23.3~mL

I hope it helps!

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Which statement defines specific heat capacity for a given sample
avanturin [10]

The statement that defines the specific heat capacity for a given sample is the quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

<h3>What is specific heat capacity?</h3>

Specific heat capacity is the of heat to increase the temperature per unit mass.

The formula to calculate the specific heat is Q = mct.

The options are attached here:

  1. The temperature of a given sample is 1 %.
  2. The temperature that a given sample can withstand.
  3. The quantity of heat that is required to raise the sample's temperature by 1 °C1 °C (Kelvin).
  4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

Thus, the correct option is 4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

Learn more about specific heat capacity

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8 0
2 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
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denis-greek [22]

Answer:

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Explanation:

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8 0
3 years ago
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katen-ka-za [31]
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3 years ago
If no carbon dioxide is present in your test tube it will be a ____________ color. If a medium amount of CO2 is present, your te
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