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4 years ago

15

Substance _ involves repeated, harmful use of a substance whereas substance _ is a physical and/or psychological addicti

on to a harmful substance.

1 answer:

alina1380 [7]4 years ago

6 0

Substance abuse... Substance addiction or dependence??

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How many moles of N are in<br> 0.227g N₂0?

klasskru [66]

Answer:

<u>0</u><u>.</u><u>2</u><u>1</u><u>7</u> ×2. =9.86×10moles

44

7 0

2 years ago

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An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its

dangina [55]

Answer:

$\frac{1}{\lambda_{6-5}} \approx 752nm$

Explanation:

From the question we are told that

Light wavelength $\lambda_l=655nm$

Light wavelength atom fall $x_L=5th-2nd$

Photon wavelength $\lambda_p=435nm$

Photon wavelength atom fall $x_P=^th-2nd$

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

$\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )$

Therefore for initial drop of 5th to 2nd

$\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )$

Therefore for initial drop of 6th to 2nd

$\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )$

Generally we subtract (5th to 2nd) from (6th to 2nd)

$\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )$

$\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )$

$\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}$

$\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}$

$\frac{1}{\lambda_{5-6}}=1.33*10^{-3}$

Therefore for 6th to 5th stage is mathematically given by

$\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}$

$\frac{1}{\lambda_{6-5}}=751.879nm$

$\frac{1}{\lambda_{6-5}} \approx 752nm$

6 0

3 years ago

By definition, any substance that accepts a proton is best defined as?

RideAnS [48]

<span> the answer is a Bronsted-Lowry base </span>

4 0

3 years ago

Which of the following are correctly paired?

mart [117]

Where’s the question tho?

8 0

3 years ago

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Water (3010 g ) is heated until it just begins to boil. If the water absorbs 5.79x105 J of heat in the

PolarNik [594]

Answer:

54 °C

Explanation:

Step 1: Given and required data

Mass of water (m): 3010 g

Heat absorbed (Q): 5.79 × 10⁵ J

Specific heat of water (c): 4.184 J/g.°C

Initial temperature (T₁): ?

Final temperature (T₂): 100 °C (Boiling point of water)

Step 2: Calculate the initial temperature of water

We will use the following expression.

Q = c × m × (T₂ - T₁)

5.79 × 10⁵ J = 4.184 J/g.°C × 3010 g × (100 °C - T₁)

T₁ = 54 °C

8 0

3 years ago