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3 years ago

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Enter your answer in the provided box. Calculate the volume of a 1.420 M NaOH solution required to titrate 40.50 mL of a 1.500 M

H3PO4 solution. 42.78 mL

1 answer:

Nikolay [14]3 years ago

7 0

Explanation:

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A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that

UkoKoshka [18]

Answer:

The correct answer is 146 g/mol

Explanation:

<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

<em>Therefore, the molar mass of the compound is 146 g/mol </em>

6 0

3 years ago

Determine the number of centiliters in 2.4 x 10^3 milliliters.

AnnZ [28]

2.4(10^3)

=2.4*10^3

=2.4*(10*10*10)

=2400 <span>milliliters

To </span>centiliters is<span> 2400mL= <u>240.0000cl. </u> </span>

8 0

3 years ago

Hanna hypothesizes that granite is denser than sandstone. How can Hanna test her hypothesis?

kherson [118]

To test if the hypothesis is correct, a good way is to think of it this way:
Density = mass/volume, right?
Calculate the mass and volume of each and do the equation; this will test your hypothesis.
You will be left with the density of each. But, make sure that the sample sizes are the same (controlled variable) otherwise it will be an unfair test.

5 0

3 years ago

8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8

garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7) kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

3 0

3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago