How much of sliver metal will weight exactly 2500.0

The atmospheric pressure on the surface of Venus is 6.84X10^4. Calculate the atmospheric pressure in atm and torr. Round each of

quester [9]

Answer:

0.675 atm

513 Torr

Explanation:

Given is that, the atmospheric pressure on the surface of Venus is

6.84 X 10⁴ Pa.

1 atm (atmospheric pressure) is equal to 101325 pascal (Pa).

To convert divide the pressure value by 101325.

Pressure in atm = $\frac{6.84 \times 10^{4} }{101325}$

= 0.675055 atm

Rounding it off to 3 significant digits: 0.675 atm

Now, one Torr is 133.322 Pa. For conversion, divide the pressure value by 133.322.

Pressure in Torr = $\frac{6.84 \times 10^{4} }{133.322}$

=513.04219 Torr

Rounding it off to 3 significant digits: 513 Torr

5 0

3 years ago

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

3 years ago

Which of the following is true if the net force on an object is zero? (1 pt) a. The object must be at rest. b. The object’s spee

prohojiy [21]

Explanation:

the object must be at rest

5 0

3 years ago

What is the average mass of an electron? A. 1 amu B. 1,837 amu C. 1/1,837

diamong [38]

Answer:

C. 1/1,837

Explanation:

3 0

3 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

3 years ago