How was the earth crust percentage determined ?

what is the molarity of a solution prepared by dissolving 1.25g of HCL gas into enough water to make 30cm cube of solution?

Tpy6a [65]

Molars are cavities in your teeth the dentists will scrape them out and fill them in for you at your earliest convinence need help with anything else while im here i can assist you!

4 0

2 years ago

500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C

dangina [55]

Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v= $\frac{1,000 g CaCl2}{500mL}$ ×100 = 0,2%w/v

b) Ratio strength is a way to express concentration. For w/v is in 1g of solute <em>r</em> mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × $\frac{1mol}{110g}$ = 9,09×10⁻³ moles of CaCl₂

500 mL of solution × $\frac{1L}{1000mL}$ = 0,500 L of solution

M = $\frac{9,09x10^{-3} moles }{0,500 L}$ = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × $\frac{0,8 kg}{1L}$ =<em> </em><em>0,625 kg of solution</em>

m = $\frac{9,09x10^{-3}moles }{0,625 kg}$ = 0,0145 m

e) In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L × $\frac{0,0182 CaCl2 moles}{1L}$ × $\frac{1equivalent}{2 moles}$ = 0,0137 equivalents

I hope it helps!

7 0

3 years ago

An elastic cord can be stretched to its elastic limit by a load of 2N.If a 35cm length of the cord is extended 0.6cm by a force

Levart [38]

Explanation:

$from \: hookes \: law \\ F = k.e \\ but \: e = 0.6 \: cm \\ 2 = k \times 0.6 \\ k = 3.33 \\ when \:F \: is \: 2.5 \\ 2.5 = 3.33 \times e {}^{.} \\ {e}^{. } = 0.75 \: cm \\ new \: length = 35 - {e}^{.} \\ = 35 - 0.75 \\ = 34.25 \: cm$

3 0

2 years ago

A sample of carbon dioxide occupies a 5.13 dm3 container at STP. What is the volume of the gas at a pressure of 286.5 kPa and a

suter [353]

Considering the ideal gas law and STP conditions, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Ideal gas law</h3>

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Volume of gas</h3>

In first place, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide, 5.13 L (5.13 dm³= 5.13 L, being 1 dm³= 1 L) are occupied by how many moles of carbon dioxide?

$amount of moles of carbon dioxide=\frac{5.13 Lx1 mole of carbon dioxide}{22.4 L}$