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Masja [62]
3 years ago
11

All 6 sides of a cube measure 65mm. Calculate the volume of the cube in cm32

Chemistry
1 answer:
zmey [24]3 years ago
8 0
Same I need help also
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1. What is a source of new alleles in natural populations?
scoray [572]

Explanation:

Generally, mating outside of a given population can give different alleles. Mutations and the crossing over of chromatids can add difference to the genetic pool.

8 0
3 years ago
Number of Mg(OH)2 formula units in 7.40 moles of Mg(OH)2.
-Dominant- [34]
The answer is 4.45 × 10²⁴ units.

To calculate this, we will use Avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance:
6.02 × 10²³ units per 1 mole

So, we need a proportion:
If 6.02 × 10²³ units are in 1 mole, how many units will be in 7.40 moles:
6.02 × 10²³ units : 1 mole = x : 7.40 moles

After crossing the products:
1 mole * x =  7.40 moles * 6.02 × 10²³ units
x = 7.40 * 6.02 × 10²³ units
x = 44.5 × 10²³ units = 4.45× 10²⁴ unit
5 0
3 years ago
What is the formula for Dicarbon trioxide
Veronika [31]

Answer:

diboron trioxide Formula: B 2 O 3 Molecular weight: 69.620 CAS Registry Number: 1303-86-2

Explanation:

<h3><u><em>hope that helps you</em></u>╰(*°▽°*)╯</h3>
4 0
3 years ago
What does it mean for energy to be conserved
marysya [2.9K]

Answer: It means that the energy is being stored or preserved to be used at a later time.

Your question was a little vague, so if this isn't the answer you were looking for, just let me know and I'll fix it.

Hope this helps! Good luck! :D

6 0
3 years ago
While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°
Lana71 [14]

Answer:

Final temperature  = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 -  25°C

305 cal = 5.25cal/°C × T2 -  25°C

305 cal / 5.25cal/°C = T2 -  25°C

58.1 °C = T2 -  25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

7 0
3 years ago
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