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Viktor [21]
3 years ago
8

The electron configuration 1s2 is correct for an element that has 2 electrons, both in the first principle energy level

Chemistry
1 answer:
Molodets [167]3 years ago
5 0

Answer:

The answer is True

Explanation:

I got it right on my quiz, hope this helps!

You might be interested in
Given the following reaction: 3CuCl2(aq) 2Na3PO4(aq) --> Cu3(PO4)2(s) 6NaCl(aq) MM (g/mol) 134.45 163.94 380.58 58.44 If 285
nata0808 [166]

Answer:

227.78g of the precipitate are produced

Explanation:

Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).

To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:

<em>Moles CuCl2:</em>

285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2

<em>Moles Cu3(PO4)2:</em>

1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2

<em>Mass Cu3(PO4)2 -380.58g/mol-</em>

0.5985 moles Cu3(PO4)2 * (380.58g/mol) =

227.78g of the precipitate are produced

7 0
3 years ago
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
2 years ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula
DerKrebs [107]

Answer:

1 kg

2 kg

3 kg

4 kg

Explanation:

i did the assignment

8 0
3 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
2 years ago
Which Statement correctly describes protons
____ [38]

Answer:

D.  

They have a positive charge and are present in the nucleus of an atom along with the neutrons.

explanation:

Protons have a positive charge.

7 0
2 years ago
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