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zepelin [54]
3 years ago
8

What is inside of a galactic center? (Inner bulge/bar of a galaxy) What kind of gases does it contain?

Chemistry
1 answer:
vredina [299]3 years ago
8 0
The galactic center, meaning the center of the milky way contains millions of stars all of various types
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Binary Covalent (molecular) Compounds
prisoha [69]

Answer:

why is ur name different sexxxy............................................................

Explanation:

5 0
2 years ago
Express in scientific notation. Remember, M must be a number such that 1 -<br><br><br> 0.000000000000013= ____
Alex

Answer:

A) 1.3 * 10^-13

Explanation:

0.000000000000013 = 1.3 * 10^-13

5 0
1 year ago
a sample of fully saturated clay weighs 1350 g in its natural state and 975 g after drying. what is the natural water content of
svetoff [14.1K]

Answer:

The soil has a percentage of water by mass of 27.8 %.

Explanation:

Keeping in mind that

  • Mass of Clay = Mass of water + Mass of Dry Soil

we can <u>calculate the mass of water</u>:

  • 1350 g = Mass of Water + 975 g
  • Mass of Water = 375 g

We can then <u>calculate the mass percentage of water in the soil</u>:

  • 375 / 1350 * 100% = 27.8 %
7 0
2 years ago
A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope.
dexar [7]
The answer is 3.

<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:

</span>(1/2) ^{n} =x

We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16

If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5% 
Presented as decimal amount:
x = 0.125


Now, let's implement this in the equation: 

<span>(1/2) ^{n} =0.125
</span>
Because of the exponent, we will log both sides of the equation:
n * log(1/2) = log(0.125)
n = \frac{log(0.125)}{log(1/2)}
<span>n = \frac{log(0.125)}{log(0.5)}
</span>n= \frac{-0.903}{-0.301}
n = 3

Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>
4 0
3 years ago
Read 2 more answers
An aluminum kettle weights 1.05 kg and has a heat capacity of 0.9211 J over grams Celsius how much heat is required to increase
Serggg [28]

Answer:

64799.4 J

Explanation:

The following data were obtained from the question:

Mass (M) = 1.05 kg = 1.05 x 1000 = 1050g

Specific heat capacity (C) = 0.9211 J/g°C

Initial temperature (T1) = 23°C

Final temperature (T2) = 90°C

Change in temperature (ΔT) = T2 – T1 =

90°C – 23°C = 67°C

Heat required (Q) =....?

The heat required to increase the temperature of the kettle can b obtain as follow:

Q = MCΔT

Q = 1050 x 0.9211 x 67

Q = 64799.4 J

Therefore, 64799.4 J of heat is required to increase th temperature of the kettle from 23°C to 90°C.

4 0
3 years ago
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