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kicyunya [14]
4 years ago
8

What is the boiling point of water and its melting point

Chemistry
2 answers:
Angelina_Jolie [31]4 years ago
4 0
The boiling point of water is 100 degrees Celsius. The melting point is 0 degrees Celsius.<span />
bazaltina [42]4 years ago
4 0
Boiling: 212 degrees F or 100 degrees Celsius, Melting: 32 degrees F or 37 degrees Celsius
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228 90th then decays by emitting alpha particles until 212 82pb is formed. how many alpha decays will occur during this chain pr
andriy [413]

Number of alpha decays is 4.

₉₀Th²³² → ₂He⁴ + ₈₈Ra²²⁸        ∴ α = ₂He⁴

Since, atomic number change by 2 in each alpha decay.

Total charge = 90 -82 = 8

Number of alpha decay = 8/2 = 4.

Alpha decay or α-decay is a type of radioactive decay wherein an atomic nucleus emits an alpha particle (helium nucleus) and thereby transforms or 'decays' into a extraordinary atomic nucleus, with a mass wide variety this is decreased through 4 and an atomic wide variety this is decreased through two.

Alpha decay - A common mode of radioactive decay wherein a nucleus emits an alpha particle (a helium-four nucleus). Beta decay - A commonplace mode of radioactive decay in which a nucleus emits beta particles. The daughter nucleus will have a better atomic number than the original nucleus.

Learn more about alpha decays here:- brainly.com/question/1898040

#SPJ4

8 0
1 year ago
Pls answer this question and have a AMAGING day!! :))
ladessa [460]

Answer:

B

Explanation:

because it is being cooled down

hoped this helps

3 0
3 years ago
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init
Jlenok [28]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

                          H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
The atomic number of an element is 29. What is the electron configuration of this element?
NNADVOKAT [17]
The answer should be C. I dont know know if you have misspellings or not
3 0
4 years ago
Read 2 more answers
What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?
rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

5 0
3 years ago
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