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2 years ago

The mass of an object is described in ________. a.meters b.cubic meters c.liters d.grams

Chemistry

1 answer:

Viefleur [7K]2 years ago

3 0

The mass of an object is describes in D. Grams.

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Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit

Aleksandr-060686 [28]

Answer:

+ axit

CH2=CH-CH2-COOH,

CH3-CH=CH-COOH (tính cả đồng phân hình học)

CH2=C(CH3)-COOH.

+ este

HCOOCH=CH-CH3 (tính cả đồng phân hình học)

HCOO-CH2-CH=CH2,

HCOOC(CH3)=CH2.

CH3COOCH=CH2

CH2=CH-COOCH3

8 0

3 years ago

If a sample of nitrogen monoxide occupies 268.0 L of space at STP, how many moles of nitrogen monoxide are

kow [346]

Answer:

1m×268.0L/22.4L=11.9m

Explanation:

there are 11.9 moles of nitrogen monoxide are present

8 0

2 years ago

15A.4(a) What is the temperature of a two-level system of energy separation equivalent to 400 cm−1 when the population of the up

maksim [4K]

Answer:

T = 525K

Explanation:

The temperature of the two-level system can be calculated using the equation of Boltzmann distribution:

$\frac{N_{i}}{N} = e^{-\Delta E/kT}$ (1)

where Ni: is the number of particles in the state i, N: is the total number of particles, ΔE: is the energy separation between the two levels, k: is the Boltzmann constant, and T: is the temperature of the system

The energy between the two levels (ΔE) is:

$\Delta E = hck$

where h: is the Planck constant, c: is the speed of light and k: is the wavenumber

$\Delta E = 6.63\cdot 10^{-34} J.s \cdot 3\cdot 10^{8}m/s \cdot 4 \cdot 10^{4}m^{-1} = 7.96 \cdot 10^{-21}J$

Solving the equation (1) for T:

$T = \frac{-\Delta E}{k \cdot Ln(N_{i}/N)}$

With Ni = N/3 and k = 1.38x10⁻²³ J/K, the temperature of the two-level system is:

$T = \frac{-7.96 \cdot 10^{-21}J}{1.38 \cdot 10^{-23} J/K \cdot Ln(N/3N)} = 525K$

I hope it helps you!

3 0

3 years ago

What is not produced during glycosis

a_sh-v [17]

The energy containing electron transporters of FADH2 are not produced during glycolysis.

4 0

3 years ago

A technical machinist is asked to build a cubical steel tank that will hold 195 L of water. Calculate in meters the smallest pos

Aleks04 [339]

Answer:

The smallest possible inside length of the tank is $0.579$ m.

Explanation:

As we know that

$1 m^3 = 1000 L$

Thus, volume of $195$ liter tank is also equal to $0.195$ cubic meter

The volume of a cube is equal to $x^3$ , where, x is the length of the side of the cube

With the give condition,

$x^ 3 = 0.195$

Solving the above equation, we get -

$x = (0.195)^{\frac{1}{3})}\\x = 0.579$

The smallest possible inside length of the tank is $0.579$ m.

4 0

3 years ago