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erica [24]
3 years ago
11

State which separation method you would use to separate potassium iodine solution​

Chemistry
1 answer:
navik [9.2K]3 years ago
3 0
Could be separated by distillation.
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How many molecules of N2 are in a 400.0 mL container at 780 mm Hg and 135°C? Avogadro’s number = 6.022 x 1023A) 7.01 × 1021 mole
ASHA 777 [7]

Answer:

B

Explanation:

Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation

PV = nRT

n = PV/RT

The parameters have the following values according to the question:

P = 780mmHg, we convert this to pascal.

760mHG = 101325pa

780mmHg = xpa

x = (780 * 101325)/760 = 103,991 Pa

V= 400ml = 0.4L

T = 135C = 135 + 273.15 = 408.15K

n = ?

R = 8314.463LPa/K.mol

Substituting these values into the equation yields the following:

n = (103991 * 0.4)/(8314.463 * 408.15)

= 0.012 moles

Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules

8 0
3 years ago
Read 2 more answers
1.Grassland are where most of our_____ comes from?
Dafna11 [192]

Answer is B.crops used as food

3 0
4 years ago
Read 2 more answers
The equilibrium reaction below has the Kc = 0.254 at 25°C. If the temperature of the system at equilibrium is decreased to 0°C,
tia_tia [17]

Answer:

The equilibrium shifts towards reagents

Explanation:

Is known Kc=\frac{[products]}{[reagents]}  rised to the power of their number of moles in the balanced reaction. When you have a system at equilibrium with Kc < 1, it means [products] < [reagents] and the system needs energy to react, so if you decrease tempeture the equilibrium shifts towards reagents and less products will be created.

This efect can be discribed with Van´t Hoff equation: ln(\frac{K1}{K2}) =-\frac{dH}{R} (\frac{1}{T1} -\frac{1}{T2} ) where we can see that if we decrease temperature (this is T2<T1) in consecuense K2<K1 and reaction doesn´t happen.

7 0
3 years ago
Calculate the volume of 12.0 g of helium at 100°c and 1.2 atm
hodyreva [135]

Volume = nRT/P

n = number of particles (moles)

R = universal gas constant (0.0821)

T = temperature (Kelvin)

P = pressure (atm)

(Assuming you have 1 mole of Helium in a chemical reaction) We would need to convert grams to moles: 12.0g He x 1 mol He/4 molar mass of He = 3 mol He

Convert Celsius to Kelvin: 100*C + 273.15 = 373.15 K

Now we can set up the equation for volume: (3mol)(0.0821)(373.15)/1.2atm = 76.6 L of Helium gas

8 0
3 years ago
2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom
lakkis [162]

<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.  

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For _{30}^{64}\textrm{Zn} isotope:</u>

Mass of _{30}^{64}\textrm{Zn} isotope = 63.929 amu

Percentage abundance of _{30}^{64}\textrm{Zn} isotope = 48.63 %

Fractional abundance of _{30}^{64}\textrm{Zn} isotope = 0.4863

  • <u>For _{30}^{66}\textrm{Zn} isotope:</u>

Mass of _{30}^{66}\textrm{Zn} isotope = 65.926 amu

Percentage abundance of _{30}^{66}\textrm{Zn} isotope = 27.90 %

Fractional abundance of _{30}^{66}\textrm{Zn} isotope = 0.2790

  • <u>For _{30}^{67}\textrm{Zn} isotope:</u>

Mass of _{30}^{67}\textrm{Zn} isotope = 66.927 amu

Percentage abundance of _{30}^{67}\textrm{Zn} isotope = 4.10 %

Fractional abundance of _{30}^{67}\textrm{Zn} isotope = 0.0410

  • <u>For _{30}^{68}\textrm{Zn} isotope:</u>

Mass of _{30}^{68}\textrm{Zn} isotope = 67.925 amu

Percentage abundance of _{30}^{68}\textrm{Zn} isotope = 18.75 %

Fractional abundance of _{30}^{68}\textrm{Zn} isotope = 0.1875

  • <u>For _{30}^{70}\textrm{Zn} isotope:</u>

Mass of _{30}^{70}\textrm{Zn} isotope = 69.925 amu

Percentage abundance of _{30}^{70}\textrm{Zn} isotope = 0.62 %

Fractional abundance of _{30}^{70}\textrm{Zn} isotope = 0.0062

Putting values in equation 1, we get:

\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]

\text{Average atomic mass of Zinc}=65.40amu

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0
3 years ago
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