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Misha Larkins [42]
3 years ago
8

Why is it important for scientists to use a classification

Chemistry
1 answer:
Kobotan [32]3 years ago
7 0
So they can tell what exact species it is. 
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I need help with my science vocabulary!!! Please help!
AVprozaik [17]

Answer:

1. B

2. E

3. A

4. D

5. C

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Hope I helped!!!

GL :)

5 0
3 years ago
how does one describe the degree of saturation in terms of polarity and amounts of solutes and solvent
Svet_ta [14]
Like dissolves Like.
7 0
3 years ago
A mass spectrometer is being used to monitor air pollutants. It is difficult, however, to separate molecules with nearly equal m
tatyana61 [14]

Answer:

4.2 m

Explanation:

The energy (E) of the selector region of the spectrometer is:

E = v*B (equation 1)

Where v is the velocity, and B the magnetic field.

The force with which particle forms a curved path can be calculated by:

\frac{mv^2}{r} = qB'

Where r is the radius of the particle, m is the mass, and q the electric charge. So:

r = \frac{mv}{qB'}

For equation 1: v = E/B, so:

r = \frac{mE}{qBB'}

The two species will be separated by Δr = 0.42 mm

Δr = Δm*E/(q*B*B')

E/(q*B*B') = r/m

Δr = Δm*r/m

r = m*Δr/Δm

r is the large of curvature, m is the avarege mass = (28.0106 + 28.0134)/2 = 28.012:

r = (28.012*0.42)/(28.0134 - 28.0106)

r = 11.76504/(0.0028)

r = 4201.8 mm

r = 4.2 m

3 0
3 years ago
A 59.0 mL portion of a 1.80 M solution is diluted to a total volume of 258 mL. A 129 mL portion of that solution is diluted by a
julia-pushkina [17]

Answer:

0.170 M

Explanation:

As this is a <em>series of dilutions</em>, we can continuosly<em> use the C₁V₁=C₂V₂ formula </em>to solve this problem:

For the first step:

  • 59.0 mL * 1.80 M = 258 mL * C₂
  • C₂ = 0.412 M

Then for when 129 mL of that 0.412 M are diluted by adding 183 mL of water:

  • V₂ = 129 mL + 183 mL = 312 mL

Using <em>C₁V₁=C₂V₂:</em>

  • 129 mL * 0.412 M = 312 mL * C₂
  • C₂ = 0.170 M
4 0
3 years ago
Part A: Calculate the mass of butane needed to produce 75.6g of carbon dioxide.
lisabon 2012 [21]

Answer:

Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.

Explanation:

Part B:

The mass of water produced when 4.86 g of butane(C4H10) react with excess oxygen is calculated as below

calculate the moles of C4H10 used = mass/molar mass

moles = 4.86g/58 g/mol =0.0838 moles

write a balanced equation for reaction

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

by use of mole ratio between C4H10 to H2O which is 2:10 the moles of

H20= 0.0838 x10/2 = 0.419 moles of H2O

mass = moles x molar mass

=0.419 molx 18 g/mol = 7.542 grams of water is formed

5 0
2 years ago
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