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4 years ago

Why are gamma rays not affected by a magnet as they Passover it

Chemistry

2 answers:

avanturin [10]4 years ago

7 0

Gamma rays are not magnetic

BartSMP [9]4 years ago

7 0

It is because they have no charge and they are waves. So, it basically means that magnet affect only charged substances, neutral substances like gamma rays, x-rays are not affected.
Hope this helps!

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A 30.5-g sample of water at 300. K is mixed with 48.5 g water at 350. K. Calculate the final temperature of the mixture assuming

ahrayia [7]

Answer:

sorry but i cant

Explanation:

8 0

3 years ago

How many chiral carbon atoms does the monosaccharide altrose have?

lina2011 [118]

4 In the open chain, 5 in the cyclic. Just like glucose.

6 0

3 years ago

Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit

djverab [1.8K]

Answer:

1.64x10⁻¹⁸ J

Explanation:

By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.

When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:

E = hc/λ

Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.

The wavelength can be calculated by:

1/λ = R*(1/nf² - 1/ni²)

Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:

1/λ = 1.097x10⁷ *(1/1² - 1/2²)

1/λ = 8.227x10⁶

λ = 1.215x10⁻⁷ m

So, the energy is:

E = (6.626x10⁻³⁴ * 3.00x10⁸)/(1.215x10⁻⁷)

E = 1.64x10⁻¹⁸ J

3 0

3 years ago

The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat

Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

Half life of lipase $t_{1/2}$ = 8 min x 60 s/min

= 480 s

Rate constant for first order reaction is as follows.

$k_{d} = \frac{0.6932}{480}$

= $1.44 \times 10^{-3}s^{-1}
$

Initial fat concentration $S_{o}$ = 45 $mol/m^{3}$

= 45 mmol/L

Rate of hydrolysis $V_m_{o}$ = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = $S_{o} (1 - X)$

= 45 (1 - 0.80)

= 9 $mol/m^{3}$

or, = 9 mmol/L

It is given that $K_{m}$ = 5mmol/L

Therefore, time taken will be calculated as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

Now, putting the given values into the above formula as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

= $-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]$

= $1642.83 s \times \frac{1 min}{60 sec}$

= 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0

3 years ago

How many moles are there in 2.75L of a 2.0M solution

gtnhenbr [62]

5.5 moles
2.0=x/2.75
2.0*2.75= x/2.75 (2.75)
5.5=x

7 0

3 years ago