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How many gallons of a 60% antifreeze solution must be mixed with 70 gallons of 25% antifreeze to get a mixture that is 50% a

Neporo4naja [7]

Answer:

The answer to your question is 175 gallons

Explanation:

Data

60% antifreeze solution volume = ? = x

25% antifreeze solution volume = 25%

Final concentration 50%

Process

0.6x + 0.25(70) = 0.5(70 + x)

Simplification

0.6x + 17.5 = 35 + 0.5x

0.6x - 0.5x = 35 - 17.5

0.1x = 17.5

x = 17.5/0.1

Result

x = 175 gallons

6 0

4 years ago

a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

Calculate the H3O+ concentration and then pH in a 0.100 M HC2H3O2 solution that is

Lana71 [14]

Answer:

Please check with the attachment posted.

Please do confirm the answer.

5 0

3 years ago

1. What coefficient of H+ balances the atoms in the following half-reaction?

lubasha [3.4K]

"4" is the one coefficient of H+ among the following choices given in the question that <span>balances the atoms in the following half-reaction. The correct option among all the options that are given in the question is the fourth option or the last option. I hope that this is the answer that has actually helped you.</span>

4 0

4 years ago

Read 2 more answers

A chemist titrates 0.200 M NaOH, strong base, with 50.00 ML of 0.100 M HCI, strong acid. How many mL of NaOH will be required to

nalin [4]

Answer:

25 mL of NaOH

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 1

Finally, we shall determine the volume of NaOH required for the reaction. This can be obtained as follow:

Molarity of the base, NaOH (M₆) = 0.2 M

Molarity of the acid, HCl (Mₐ) = 0.1 M

Volume of the acid, HCl (Vₐ) = 50 mL

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 1

Volume of the base, NaOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ / n₆

0.1 × 50 / 0.2 × V₆ = 1

5 / 0.2 × V₆ = 1

Cross multiply

5 = 0.2 × V₆

Divide both side by 0.2

V₆ = 5/0.2

V₆ = 25 mL

Thus, 25 mL of NaOH is needed for the reaction

8 0

3 years ago