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Lemur [1.5K]
2 years ago
14

) Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer

n. Round your answer to the nearest integer.
Chemistry
1 answer:
RSB [31]2 years ago
4 0

Answer:

2

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

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A collection of coins contains 14 pennies, 16 dimes, and 7 quarters. What is the percentage of pennies in the collection?A colle
marin [14]

Answer:

=37.83783784

Explanation:

Find the total sum of all coins,

which is 37, take the number of pennies and the total of all coins put in parenthesis( 14/37) like so and than * times them by 100

you equation should look like this

(14/37)* 100= and than the answer shown above should be the one you received. I have checked this with multiple calculators, it should be accurate.  

8 0
2 years ago
Read 2 more answers
495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F
I am Lyosha [343]

Answer:

<u><em>Total pressure of the flask is 2.8999 atm.</em></u>

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

                                              = 0.495 L (1 cm³ = 1 mL = 0.001 L)                                            

Volume of nitrogen (N2) gas =  877 cm3

                                               = 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

                         = 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature =  25 °C

T = (25°C + 273.15) K

    = 298.15 K

Pressure = 114.7 kPa

               = 114.700 Pa

Pressure (torr) = 114,700 / 101325

                        = 1.132 atm

Formula:

PV=nRT  <em>(ideal gas equation)</em>

P = pressure

V = volume

R (gas constnt)=  0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

<u>For Oxygen:</u>

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

  = 0.560 / 24.47

  = 0.0229 moles

<u>For Nitrogen:</u>

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

  = 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

  = 0.0229 moles + 0.0406 moles

n  = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K  /  0.536 L

P = 1.554 / 0.536

<u><em>P = 2.8999 atm</em></u>

Total pressure in the flask is  2.8999 atm, while assuming the temperature constant.

7 0
2 years ago
Read 2 more answers
A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with
andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g

Molar mass of nitrogen gas, N_2 = 28.014 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass}{28.014\ g/mol}

Mass of nitrogen gas = 28.014x g

So,

Let, Mass_{Container}=y

y+28.014x=37.289

Similarly,

Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g

Molar mass of nitrogen gas, CO_2 = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass{44.01\ g/mol}

Mass of nitrogen gas = 44.01x g

So,

y+44.01x=37.440

Solving the two equations, we get :

Mass_{Container}=y=37.025\ g

x = 0.00943 moles

Thus, Given:

Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g

37.025\ g+Mass_{Unknown\ gas}=37.062\ g

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00943\ moles= \frac{0.037\ g}Molar mass}

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0
2 years ago
Consider a culture medium on which only gram-positive organisms such as Staphylococcus aureus colonies can grow due to an elevat
Olin [163]
<h2>Selective & Differential Medium</h2>

Explanation:

  • Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
  • Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
  • Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria

4 0
2 years ago
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
2 years ago
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