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Lorico [155]
3 years ago
9

What is the mass of 2.47 moles of carbon dioxide CO2 ?

Physics
2 answers:
Anna11 [10]3 years ago
6 0

Answer:

CO2 molecule is made up of Carbon (atomic mass =12) and oxygen (atomic mass=16).

So first finding the mass of 1 molecule of CO2 which is equals to

= mass of 1 carbon atom + masses of 2 oxygen atom, we get

= 12+(16*2)= 12+32= 44 a.m.u.

Now 1 molecule of CO2 has mass 44 amu so mass of 1 mole CO2 will be 44 grams.( 1 a.m.u.=1.6729*10^-33 grams. 1 mole = 6.022*10^23, so 44 a.m.u.=73.6076*10^-33 grams approx. For one mole CO2, 73.6076*10^-33*6.022*10^23 which is approximately equals to 44 grams. )

1 mole CO2= 44grams, so 2.5 moles = 44*2.5= 110 grams

So our answer is 110 grams

BARSIC [14]3 years ago
5 0

Answer: yall its 0.110

Explanation: trust me lol

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After hearing about an accident on his normal route, Mr. Gujral checks for alternate routes to get to work. What type of circuit
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Answer:

C

Explanation:

a parallel circuit because there is more than one path

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3 years ago
Situation A bichromatic source produces light having wavelengths in vacuum of 450 nm and 650 nm. The indices of refraction are 1
erica [24]

Answer:

\theta=0.52^{\circ}

Explanation:

It is given that,

Wavelength in vacuum, \lambda_1=450\ nm

Wavelength in vacuum, \lambda_2=650\ nm

Refractive index for air, n_1=1

First refractive index, n =1.44

Second refractive index, n' =1.42

A ray is incident at an angle of incidence of 50 degrees. Let r is the angle of refraction. Firstly calculating the angle of refraction for two values of wavelength from Snell's law as :

\dfrac{sin\ i}{sin\ r}=\dfrac{n_2}{n_1}

r=sin^{-1}(\dfrac{n_1\ sin\ i}{n})

For 450 nm, r=sin^{-1}(\dfrac{1\ sin(50)}{1.44})

r = 32.13 degrees

For 650 nm, r'=sin^{-1}(\dfrac{1\ sin(50)}{1.42})

r' = 32.65 degrees

Let \theta is the angle of dispersion between the two refracted rays in the oil such that,

\theta=r'-r

\theta=32.65-32.13

\theta=0.52^{\circ}

So, the angle of dispersion between the two refracted rays in the oil is closest to 0.52 degrees.

3 0
3 years ago
A bicycle rider travels 50.0 Km in 2.5 hours. What is the cyclist's average speed?
vekshin1
<span>If you divide the numbers you will get on average of 20km</span>
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4 years ago
Read 2 more answers
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
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What does the n stand for? Is it north?
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