Briefly describe the intermolecular forces present in your unknown

1 answer:

7 0

An inter-molecular power is basically an alluring power between neighboring particles. There are three regular sorts of inter-molecular power: lasting dipole-dipole powers, hydrogen bonds and van der Waals' powers.

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WILL MARK BRAINIEST Can someone please help me understand this?

Mila [183]

To convert from moles to grams you divide by the molar mass of the element. To convert from grams to moles you X by the molar mass element

5 0

3 years ago

4.00 g of O2 gas are in a sealed, 2.00 L gas canister at 22.0 °C what is the pressure inside this container (in atm)?

Volgvan

Answer:

1.51448 atms

Explanation:

6 0

2 years ago

(You do) If you have 47.2 mol of Na available, along with an excess of Cl₂, how many grams of NaCl can you produce?

IrinaVladis [17]

Answer:

2,760 grams NaCl

Explanation:

To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.

2 Na + Cl₂ --> 2 NaCl

Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol

Molar Mass (NaCl) = 58.44 g/mol

47.2 moles Na 2 moles NaCl 58.44 grams

---------------------- x --------------------------- x ------------------------- =
2 moles Na 1 mole NaCl

= 2,758.368 grams NaCl

= 2,760 grams NaCl

5 0

1 year ago

What is the formula of this inorganic salt hydrate?

barxatty [35]

Salt hydrates are an important class of PCMs. An inorganic salt hydrate (hydrated salt or hydrate) is an ionic compound in which the ions attract a number of water molecules, which are then trapped inside the crystal lattice. A hydrated salt has the generic formula MxNy. nH2O.

7 0

2 years ago

The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse

Umnica [9.8K]

Answer : 121.5 μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

 $\frac{N}{ N_{0} } = e^{-( \frac{0.693 X T_{2} }{T_{1}})$

3.8 / $N_{0}$ = $e^$ ((0.693 X 162.5 ) / 32.5) = 121.5

On solving we get, The initial activity as 121.5 μci

5 0

2 years ago

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