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3 years ago

How could gamma rays,for example change something irreversibly

Chemistry

2 answers:

ANTONII [103]3 years ago

6 0

compute the average rate of change of the function f(x)=(the square root of)4x+8 as it moves from x1=-2 to x2=7

coldgirl [10]3 years ago

5 0

They emit radiation. Which can be very harmful to living things, as well as some non living things. I hope i answered you question :)

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Carbon dioxide is released into the atmosphere by _____.

kirill115 [55]

Answer:

4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;

Explanation:

a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation

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2 years ago

Will give most brainly.

Anna [14]

Answer:

One mole

Explanation:

It tells u the concentration

8 0

2 years ago

The heisenberg uncertainty principle status that

Juli2301 [7.4K]

The Heisenberg uncertainty principle listed above states that it’s impossible to determine with high precision both the momentum and position of an electron simultaneously.

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2 years ago

How many moles are in 4.01x10^24 particles?

skelet666 [1.2K]

Answer:

6.64

Explanation:

so moles

$\frac{number \: of \: \: particles \: }{6.02 \times {10}^{23} }$

then do =4.01×10^24/6.02×10^23

you get the answer as 6.6445 u only need to write the answer to

3.S.F 6.64 moles

hope this make sense :)

6 0

3 years ago

If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas mol

Free_Kalibri [48]

Answer:

The root mean square speed of O2 gas molecules is

519.01 m/s

Explanation:

The root mean square velocity :

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$K.E_{avg}=\frac{3}{2}RT$

$K.E =\frac{1}{2}mv_{rms}^{2}$

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =

$K.E_{avg}=\frac{3}{2}RT$

$T=\frac{2(K.E)}{3(R)}$

K.E of He = 4310 J/mol

$T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}$

$T=345.60K$

Now , Use Vrms to calculate the velocity of O2

$v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}$

$v_{rms}=\sqrt{\frac{8619.9552}{0.032}}$

$v_{rms}=\sqrt{26935.001}$

$v_{rms}=519.01m/s$

6 0

3 years ago