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3 years ago

Anyone know the answer to this one?

Chemistry

1 answer:

sergey [27]3 years ago

5 0

Answer:

C) 0.24 M

Explanation:

The given chemical reaction is presented as follows;

H₂SO₄(aq) + 2 KOH(aq) → K₂SO₄(aq) + 2 H₂O(l)

The titration experiment results are;

Volume of H₂SO₄(aq) used = 12.0 mL

Volume of KOH (aq) used = 36.0 mL

Concentration of KOH (aq) = 0.16 M

The number of moles of KOH present, n = 0.16 M × 36/1000 = 0.00576 moles

From the given reaction, 1 mole of H₂SO₄ reacts with 2 moles of KOH to give 1 mole of K₂SO₄ and 2 moles of H₂O

Therefore, 0.00576 moles of KOH reacts with (1/2) × 0.00576 moles = 0.00288 moles of H₂SO₄

Therefore, for the reaction;

The number of moles of H₂SO₄ in 12.0 mL of H₂SO₄ = 0.00288 moles

The concentration of the H₂SO₄ = 0.00288 M/(12.0 mL) = 0.24 M

The concentration of H₂SO₄ in the reaction = 0.24 M.

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When a strip of magnesium metal is placed in an aqueous solution of copper(II) nitrate, elemental copper coats the surface of th

Lera25 [3.4K]

Answer:

1. 2+ ( $Cu^{2+}$ ).

2. 0 ( $Cu^0$ ).

Explanation:

Hello,

In this case, the described chemical reaction is a redox reaction in fact, since the oxidation states of both magnesium and copper change as shown due to the displacement:

$Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)$

Therefore:

1. Since copper is the cation in the copper (II) nitrate, the (II) means that its charge is 2+ ( $Cu^{2+}$ ).

2. Since copper is alone, it means no electrons are being neither shared not given, its charge is 0 ( $Cu^0$ ).

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4 0

3 years ago

he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c

Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer: The rate constant at 324°C is $61.29M^{-1}s^{-1}$

Explanation:

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{244^oC}$ = equilibrium constant at 244°C = $6.7M^{-1}s^{-1}$

$K_{324^oC}$ = equilibrium constant at 324°C = ?

$E_a$ = Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $244^oC=[273+244]K=517K$

$T_2$ = final temperature = $324^oC=[273+324]K=597K$

Putting values in above equation, we get:

$\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}$

Hence, the rate constant at 324°C is $61.29M^{-1}s^{-1}$

8 0

3 years ago

What are the relationships between temperature and viscosity of water?

iris [78.8K]

Explanation:

Both cohesion and molecular interchange contribute to liquid viscosity. The impact of increasing the temperature of a liquid is to reduce the cohesive forces while simultaneously increasing the rate of molecular interchange. The former effect causes a decrease in the shear stress while the latter causes it to increase.

temperature?

The viscosity of liquids decreases rapidly with an increase in temperature, and the viscosity of gases increases with an increase in temperature. Thus, upon heating, liquids flow more easily, whereas gases flow more sluggishly.

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8 0

3 years ago

In the laboratory you dissolve 13.9 g of potassium phosphatein a volumetric flask and add water to a total volume of 250mL.

tensa zangetsu [6.8K]

Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

Concentration of the phosphate anion? 0,262 M.

Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ (1)

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

There are 250 mL of solution≡0,25 L

The moles of K₃PO₄ are:

13,9 g of K₃PO₄ × $\frac{1mol}{212,27 g}$ = 0,0655 moles of K₃PO₄

The molarity of the solution is:

$\frac{0,0655 moles}{0,25L}$ = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

$\frac{0,1965 moles}{0,25L}$ = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

I hope it helps!

4 0

3 years ago

How many grams (m) of glucose are in 165 ml of a 5.50% (m/v) glucose solution? express your answer numerically in grams?

Leya [2.2K]

mass of glucose = 0.055 *165 = 9.075 g

vol of methyl alc = 0.185 * 1.87 = 0.346 L = 346 ml

% NaCl ( m/v ) = mass NaCl * 100/ vol of soln

or Vol of Soln = mass NaCl / % NaCl (m/v)

= 32.1 * 100 / 6 = 535 ml the total vol of soln

6 0

4 years ago