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3 years ago

An object has a weight of 5.10 Newtons (N) at sea level, where the acceleration due to gravity is 9.81 m/s2.

Chemistry

2 answers:

RUDIKE [14]3 years ago

8 0

Well, you could get the mass as $m=\frac{G_0}{g_0}$ ad then $g_1=\frac{G_1}{m}$ , where $G_0$ is the sea level weight, $g_0$ the sea level accel., $g_1$ the accel. above while $G_1$ the weight above.

Sergio039 [100]3 years ago

3 0

Answer:

Well, you could get the mass as m=\frac{G_0}{g_0} ad then g_1=\frac{G_1}{m}, where G_0 is the sea level weight, g_0 the sea level accel., g_1 the accel. above while G_1 the weight above

Explanation:

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Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is

Strike441 [17]

Answer:

For a: The molarity of commercial HCl solution is 12.39 M.

For b: The molality of commercial HCl solution is 16.79 m.

For c: The volume of commercial HCl solution needed is 2.42 L.

Explanation:

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

For a:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

For b:

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

For c:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

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