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Montano1993 [528]
2 years ago
5

How does the oceanic crust compare to continental crust?

Chemistry
1 answer:
dimulka [17.4K]2 years ago
5 0

I think the answer is number (4)

You might be interested in
pesticides and fertilizers can help farmers to produce more crops. However, overuse of these chemicals can result in?
nadezda [96]
Overuse of the same chemicals can result in the pest becoming immune to the pesticides. 
8 0
3 years ago
What is the Molar Mass of Cs2SO3?​
Airida [17]

Answer:

345.89 g/mol

Explanation:

To find the molar mass, find the atomic mass of all the elements from a periodic table.

Cs - 132.91 × 2 = 265.82

S - 32.07

O - 16.00 × 3 = 48.00

Now add them all together.

265.82 + 32.07 + 48.00 = 345.89 g/mol

Hope that helps.

8 0
2 years ago
There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a
Vanyuwa [196]
<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

  • The natural isotopic abundance of 10B is 19.60%.
  • The natural isotopic abundance of 11B is 80.40%.
  • The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

∴ x = 0.2

To get percentages  we need to multiply it by 100

So, the calculated abundance is 80% for 11 B and 20% 10  B.

5 0
3 years ago
A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24
SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0
3 years ago
The density of ethanol is 0.788 g/mL. What is the mass of 157 mL of alcohol?
Lina20 [59]

Answer:

<h3>The answer is 123.72 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 157 mL

density = 0.788 g/mL

We have

mass = 0.788 × 157 = 123.716

We have the final answer as

<h3>123.72 g</h3>

Hope this helps you

6 0
2 years ago
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