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3 years ago

How many moles are in 2.5g of n2

Chemistry

2 answers:

exis [7]3 years ago

7 0

n= m/M = 2.5/14 = 5/28 mol

ok done. thank to me :>

Thepotemich [5.8K]3 years ago

6 0

Answer:

0.089

Explanation:

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A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element pre

nordsb [41]

Answer:

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

6 0

4 years ago

Read 2 more answers

If you mix cu2+ with? a. nacl, b. nano2, c. h2o , arrange the solutions based on their absorption from highest frequency to lowe

erica [24]

The arrangement of the solutions based on their absorption from highest frequency to lowest frequency :

b. $NaN$ $O_{2}$ > c. $H_{2} O$ > a.NaCl

<h3>What is absorption frequency?</h3>

The frequency of the molecular vibration that led to the absorption is the same as the absorption frequency of a basic IR absorption band.

In a way, an emission spectrum is the opposite of an absorption spectrum.
The discrepancies in the energy levels of each chemical element's orbitals correspond to absorption lines for each chemical element at various particular wavelengths.
Therefore, it is possible to identify the constituents in a gas or liquid using its absorption spectrum.
Absorption spectroscopy is most frequently used to measure infrared, atomic, visible, ultraviolet (UV), and x-ray waves.

Learn more about Absorption frequency here:

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1 year ago

6 × 10^9 mm = m. how to convert millimeters to m

OLga [1]

Answer:

6x10^9x10^-3m=6x10^6m

Explanation:

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4 years ago

If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera

Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0

3 years ago

How do I study Chemistry efficiently?

TEA [102]

Review and Study Material Before Going to
Class.

Seek Understanding.

Take Good Notes.

Practice Daily.

Take Advantage of Lab Time.

Use Flashcards.

Use Study Groups.

Break Large Tasks Into Smaller Ones.

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3 years ago