CHEMISTRY COLLEGE

carbon dioxide at 25 degrees celsius and 101.3 kpa has a density of 1.799 kg/m^3 determine gas constant

Answers

Answer 1
Answer:

Answer:

Explanation:

Temperature of gas in absolute scale T = 25 + 273 = 298 .

pressure of gas P = 1.013 x 10⁵ N / m²

density D = 1.799 kg / m³

= 1799 g / m³

From gas formula

PV / T = n R Where P is pressure , V is volume and T is absolute temperature , n is no of moles

P / T = n R / V

P / T = m R /M V where m is mass of gas and M is molecular weight .

m / V = D ( density )

P / T = DR/ M  

PM / DT = R

Putting the values

1.013 x 10⁵ x 44  /  (1799 x 298)

R = 8.314.09 J / K mole

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1. On the basis of your results, what is the relationship between the temperature of the solvent and the rate of solution formation?

Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4

Answers

Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C, -161.5 °C  and -87.7 °C, clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

Calculate the molality of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4) (35.4 % means 35.4 g of H3PO4in 100 g of solution)

Answers

Answer:

3.6124 m/kg

Explanation:

Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us  35.4 g of H3PO4 and 100 g of solution to work with.

To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.

Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:

35.4 g x 1 mol / 97.9952 g = 0.3612422 mol

Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.

To find the molality, divide the moles of solute and kilograms of solution.

0.3612422 mol / 0.1 kg = 3.6124 m/kg

A chemical equation is balanced when thesubscripts of the reactants equal the subscripts of the products,
coefficients of the reactants equal the coefficients of the products,
products and reactants are the same chemicals,
same number of each kind of atom appears in the reactants and in the products,

Answers

same number of each kind of atom appears in the reactants and in the products

According to the valence bond theory the triple bond in ethyne consists of

Answers

Answer:
            According to the valence bond theory the triple bond in ethyne consists of one sigma bond and two pi bonds.

Explanation:
                   Atomic number of carbon is 6. The ground state electronic configuration of carbon is as follow,

                                         1s², 2s², 2p²

And the excited state electronic configuration of carbon is as follow,

                                         1s², 2s¹, 2px¹, 2py¹, 2pz¹

In ethyne the 2s¹ orbital and 2px¹ orbitals having unpaired electrons form sigma bonds by head to head overlapping with orbitals of hydrogen atom and carbon atom. The remaining 2py¹ and 2pz¹ orbitals of both carbons overlap perpendicular to the existing sigma bond resulting in the formation of two pi bonds.

Carbon dioxide and an unknown gas start to effuse from a container and the carbon dioxide takes 4.69 times as long to escape as the unknown gas. What is the identity of the unknown gas?

Answers

Answer: the gas is lighter than CO2 and it is hydrogen

Explanation:Please see attachment for explanation

4.What volume of 0.120 M HNO3(aq) is needed to
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL
​

Answers

Answer:

B) 125 mL

Explanation:

M1V1=M2V2

(0.120M)(x)=(150.0 mL)(0.100M)

x= 125 mL

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