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Salsk061 [2.6K]
3 years ago
12

The goal of this experiment was to answer the

Chemistry
1 answer:
Tanzania [10]3 years ago
3 0

Answer:

hypothesis is correct, if temp increases particle velocity increases, increasing force and number of collisions therefore, volume increases

Explanation:

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The ability of an atom in a molecule to attract electrons is best quantified by the ________.
nevsk [136]

Answer:

b. electronegativity

Explanation:

The ability of an atom in a molecule to attract electrons is best quantified by the _Electronegativity_.

Electronegativity is the ability of an atom to attract a shared pair of electron which is loosely held. Polar molecules often possess such properties. The Halogen group (group 17) is famous for such property because it has the highest electronegativity in periodic table.  

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2 years ago
2. Electron A falls from energy level X to energy level Y and releases blue light. Electron B falls from energy level Y to energ
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Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai
Furkat [3]

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Explanation:

check the attachment below

5 0
2 years ago
The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?
Nat2105 [25]
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
6 0
3 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
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