How many grams of the parent isotope are left in the sample after three half lives?

When naming bases, you add ___ to the end

dangina [55]

Answer:

hydroxide ion

Explanation:

because base produce hydroxide ion when dissolved in water

6 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

Write a ground state electron configuration for each neutral atom

Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0

2 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation: