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Phoenix [80]
2 years ago
14

A student has a 1g sample of each of the following compounds: NaCl, KBr, and KCL Which of the following lists the samples in ord

er of increasing number of moles in the sample?
Chemistry
1 answer:
krek1111 [17]2 years ago
4 0

Given :

A student has a 1 g sample of each of the following compounds: NaCl, KBr, and KCl.

To Find :

The samples in order of increasing number of moles in the sample.

Solution :

Molecular mass of NaCl, KBr, and KCl is 58.5 g/mol , 119 g/mol and

74.5 g/mol respectively .

Moles of NaCl , n_1=\dfrac{1}{58.5}=0.017\ mol.

Moles of KBr , n_2=\dfrac{1}{119}=0.008\ mol.

Moles of KCl , n_2=\dfrac{1}{74.5}=0.013\ mol.

The order of moles in increasing order is :

KBr , KCl and NaCl .

Hence , this is the required solution .

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Using the solubility curve, what is the effect of increased temperature on the solubility of KBr in 100 grams of water? The solu
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The graph is needed to answer this question.

Solubility may increase or decrease with temperature depending on the properties of the solute and the solvent.

It is quite common that the solubility of the ionic compounds, like KBr, in water increases with temperature.

Use your solubility curve for the KBr and you wiil see a line that starts at a solubility a little greater than 50 grams of the salt in 100 grams of water for temperaute 0°C and increase linearly until almost 100 grams of the salt in 100 grams of water at 100°C.

So, in this case you can affirm that the solubility of KBr increases with the temperature.

Answer: the second option: the solubility increases.
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A planet travels in an elliptical path around a star, as shown in the figure. As the planet gets closer to the star, the gravita
Soloha48 [4]

Answer:

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.

Explanation:

Force between two objects can be expressed by an equation:

F = G • m1 • m2 / r^2,

where m1 and m2 are objects' masses, r is the distance between them, and G is a gravitational constant.

That means that greater the masses or lesser the distance, the force will be greater, and vice versa.

This force exists between any two objects, but is generally extremely weak, so it's best observed with big and large objects with great mass, such as planets and stars.

This force, whatever its magnitude may be, always works on both objects, following the third Newton's law.

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3 0
2 years ago
What is the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I? (Kb of CH₃NH₂ is 4.4 × 10⁻⁴)
Karolina [17]

Answer:

\boxed{\text{10.84}}

Explanation:

A solution of a weak base and its conjugate acid is a buffer.

The equation for the equilibrium is

\rm CH$_3$NH$_2$ + H$_2$O $\, \rightleftharpoons \,$ CH$_3$NH$_2$+ H$_{3}$O$^{+}$\\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H$_2$O $\longrightarrow \,$ BH$^{+}$ + OH$^{-}$; $K_{\text{b}}$ = 4.4 \times 10^{-4}$

The Henderson-Hasselbalch equation for a basic buffer is

\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}

Data:

   [B] = 0.400 mol·L⁻¹

[BH⁺] = 0.250 mol·L⁻¹

    Kb = 4.4 × 10⁻⁴

Calculations:

(a) Calculate pKb

pKb = -log(4.4× 10⁻⁴)  = 3.36

(b) Calculate the pH

\text{pOH} = 3.36 + \log \dfrac{0.250}{0.400} = 3.36 + \log 0.625 = 3.36 - 0.204 = 3.16\\\\\text{pH} =14.00 -3.16 = \mathbf{10.84}\\\\\text{The pH of the solution is }\boxed{\textbf{10.84}}

4 0
2 years ago
A piece of aluminum foil 1.00cm square and 0.590mm thick is allowed to react with bromine to form aluminum bromide. Part A How m
nalin [4]

Answer:

Part A: 5.899x10^-3 moles of Al

Part B: 1.573 g of AlBr3

Explanation:

Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.

0.590 mm  (\frac{1 cm}{10 mm}) = 0.059 cm

V= lxlxl= (1cm)(1cm)(0.059 cm)= 0.059 cm^3.

0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:

0.059 cm^3 Al(\frac{2.699 g Al}{1 cm^3 Al})(\frac{1 mol Al}{27 g Al})= 5.899x10^-3 moles

Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:

                                               2Al + 3Br2 → 2AlBr3

We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:

5.898 x 10^-3 moles Al (\frac{2 moles AlBr3}{2 moles Al}) (\frac{266.7 g AlBr3}{1 mol AlBr3}) = 1.573 g of Al

5 0
2 years ago
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