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zavuch27 [327]
2 years ago
10

a titanium bicycle frame displays 0.250 L of water and has a mass of 1.21kg. what is the density of the titanium on g/cm3?

Chemistry
1 answer:
fomenos2 years ago
3 0

Answer:

$4.49 g/cm3

Explanation:

Density=Mass/Volume

plzz put brainiest

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A mixture of benzene and toluene has a total vapor pressure at 25°C of 45.06 mmHg. What is the partial pressure of benzene in th
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8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]
yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡  ΔH_{f} = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8  kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5  kJ/mol

Equation 4 ≡ BE(Cl2) = 226  kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332  kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

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3 years ago
How many valence electrons are in an atom of phosphorus? (atomic number 15) (1 point)?
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Gas law problem: A gas has a volume of 800 mL at -25℃ and 600 kPa. What would the volume of the gas be at 227℃ and 1000 kPa?
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<h3><u>Answer;</u></h3>

= 930.23 mL

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Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and

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Thus;

V2 = P1V1T2/T1P2

     = (600 ×800 ×500) / (258 × 1000)

     = 930.23 mL

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