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photoshop1234 [79]
3 years ago
13

3 moles of C4H10 weighs?

Chemistry
1 answer:
seraphim [82]3 years ago
7 0
From moles to grams C4H10 weighs 174.3666grams
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Josie observed that when water is boiled in an open beaker it completely disappears, leaving the beaker dry. What can Josie conc
-BARSIC- [3]
That when water is boiled in a open beaker and it disappears that it evaporates into the air
6 0
4 years ago
If I want to accelerate a mass of 3 kg at 5 m/s2 then how much force should I apply?
Stolb23 [73]

<u>Answer:</u> The force that must be applied is 15 N.

<u>Explanation:</u>

Force exerted on the object is defined as the product of mass of the object and the acceleration of the object.

Mathematically,

F=m\times a

where,

F = force exerted = ?

m = mass of the object = 3 kg

a = acceleration of the object = 5m/s^2

Putting values in above equation, we get:

F=3kg\times 5m/s^2=15N

Hence, the force that must be applied is 15 N.

5 0
3 years ago
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of
harkovskaia [24]

Answer:

The frequency of this light is 6.98\times 10^{14} s^{-1}.

Explanation:

Wavelength of the light = \lambda = 430 nm=4.30\times 10^{-7} m

Speed of the light = c = = 3\times 10^8 m/s

Frequency of the light = \nu

\nu =\frac{c}{\lambda }

\nu =\frac{3\times 10^8 m/s}{4.30\times 10^{-7} m}=6.98\times 10^{14} s^{-1}

The frequency of this light is 6.98\times 10^{14} s^{-1}.

5 0
3 years ago
Which sample contains a total of 3.0 x 10^23 molecules?
belka [17]
Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:

A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.

Therefore the only choice that fits is A. 80 g of Br2.
3 0
3 years ago
Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t
stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

4 0
3 years ago
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