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serious [3.7K]
3 years ago
6

What are chartistics of neutrons?

Chemistry
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

Neutral subatomic particle that is a constituent of every atomic nucleus except ordinary hydrogen.

Explanation:

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What is the mass of 2.49 cm of aluminum? The density of aluminum is 2.70 g/cm.
OverLord2011 [107]

Answer:

6.723

Explanation:

2.49*2.70

5 0
3 years ago
¿Que es la energía térmica?
qaws [65]
La energía térmica es la energía que proviene del calor.

De nada ;)
7 0
3 years ago
A beaker has a volume of .550L. What is it in milliliters?
Vladimir [108]
Add 3 zeros at the end to convert to milliliters.
550,000 milliliters.
6 0
3 years ago
Barbiturates are synthetic drugs used as sedatives and hypnotics. Barbital ( = 184.2 g/mol) is one of the simplest of these drug
DanielleElmas [232]

Answer:

118.75°C is the boiling point of a solution.

Explanation:

Mass of the solute that is barbiturates = 42.5 g

Molar mass of a solute = 184.2 g/mol

Moles of solute = \frac{42.5 g}{184.2 g/mol}=0.2307 mol

Mass of the solvent that acetic acid = 825 g = 0.825 kg

molality=\frac{\text{Moles of solute}}{\text{Mass of solvent}}

Molality of the solution (m):

m=\frac{0.2307 mol}{0.825 kg}=0.2796 m

Elevation in boiling point is given as:

\Delta T_b=i\times K_b\times m

i = 1 (organic compound)

=1\times 3.07^oC/m\times 0.2796 m=0.8585^oC

\Delta T_b=T_b-T

T_b = Boiling temperature of solution.

T = boiling temperature of solvent that is acetic acid=117.9°C

0.8585^oC=T_b-117.9^oC

T_b=118.75 ^oC

118.75°C is the boiling point of a solution.

6 0
3 years ago
How many moles are in 2.8x10^23 atoms of Calcium?
alina1380 [7]

Answer:

b. 0.47 moles Ca.

Explanation:

Hello there!

In this case, since 1 mole of any element contains 6.022x10²³ atoms of the same, it is possible for us to compute the moles in 2.8x10²³ atoms of calcium via the Avogadro's number:

2.8x10^{23}atomsCa*\frac{1molCa}{6.022x10^{23}atomsCa}\\\\=0.47molCa

Therefore, the answer would be b. 0.47 moles Ca.

Best regards!

5 0
3 years ago
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