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harina [27]
3 years ago
10

Determine what’s wrong with these electronic configurations

Chemistry
1 answer:
Anika [276]3 years ago
7 0

Answer:

Explanation:

19) it is 3d10 instead of 4d10

20) it is missing 3p6, and 4s2 before 3d5

21) Ra is not a noble gas

22) Cs is not a noble gas

You might be interested in
If the mass of a block of wood is 4.6g and the density is 0.98g/mL what is the volume.
irina [24]

Answer:

The answer is

<h2>4.69 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass of wood = 4.6 g

density = 0.98 g/mL

It's volume is

volume =  \frac{4.6}{0.98}  \\  = 4.69387755...

We have the final answer as

<h3>4.69 mL</h3>

Hope this helps you

3 0
4 years ago
How many grams of acetylene are produced by adding 3 moles of cac2
jolli1 [7]

Answer:

78.12g of acetylene

Explanation:

Acetylene and calcium hydroxide are produced when H₂O reacts with CaC₂. The reaction is:

2 H₂O + CaC₂ → C₂H₂ + Ca(OH)₂

<em>Where 1 mole of C₂H₂ (acetylene) is produced per mole of CaC₂</em>

Thus, the addition of 3 moles of CaC₂ produces 3 moles of acetylene.

As molecular mass of acetylene is 26.04g/mol, grams of acetylene produced are:

3mol C₂H₂ × (26.04g / mol) = <em>78.12g of acetylene</em>

8 0
3 years ago
In a test of an automobile engine 1.00 L of octane (702 g) is burned, but only 1.84 kg of carbon dioxide is produced. What is th
Reptile [31]

Answer:

The % yield of CO2 is 85.05 %

Explanation:

Step 1: Data given

Mass of octane = 702 grams

Molar mass octane = 114.23 g/mol

Mass CO2 =1.84 kg = 1840 grams

Molar mass of CO2

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of octane

Moles octane = mass octane / molar mass octane

Moles octane = 702.0 grams / 114.23 g/mol

Moles octane = 6.145 moles

Step 4: Calculate moles of CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.145 moles octane we'll have 8*6.145 moles =49.16 moles

Step 5: Calculate mass of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 49.16 moles * 44.01 g/mol

Mass CO2 = 2163.5 grams

Step 6: Calculate % yield of carbon dioxide

% yield = (actual yield / theoretical yield)*100%

% yield = (1840/2163.5)*100%

% yield = 85.05 %

The % yield of CO2 is 85.05 %

4 0
3 years ago
calculate the number of grams of CH3COONa * 3H2O (sodium acetate tri-hydrate) needed to make 250.0 mL of a CH3COOH (acetic acid)
anastassius [24]

Answer:

10.88 g

Explanation:

We have:                            

[CH₃COOH] = 0.10 M

pH = 5.25

Ka = 1.80x10⁻⁵      

V = 250.0 mL = 0.250 L

M_{CH_{3}COONa*3H_{2}O} = molar \thinspace mass = 136 g/mol

The pH of the buffer solution is:  

pH = pKa + log(\frac{[CH_{3}COONa*3H_{2}O]}{[CH_{3}COOH]})    (1)

By solving equation (1) for [CH₃COONa*3H₂O] we have:  

log [CH_{3}COONa*3H_{2}O] = pH - pKa + log [CH_{3}COOH]  

log [CH_{3}COONa*3H_{2}O] = 5.25 - (-log(1.80 \cdot 10^{-5})) + log (0.10) = -0.495                                

[CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M

Hence, the mass of the sodium acetate tri-hydrate is:

m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g

Therefore, the number of grams of CH₃COONa*3H₂O needed to make an acetic acid/sodium acetate tri-hydrate buffer solution is 10.88 g.

I hope it helps you!      

4 0
3 years ago
Analyisis Of Compound (%)<br> C=72.15 H=7.08 N=4.68 O=16.03<br> What is it?
klemol [59]

Answer:

Explanation:

Ratio of C : H : N : O  in grams

=  72.15 : 7.08 : 4.68 : 16.03 in grams

Ratio of C : H : N : O in moles

72.15/12 : 7.08/1 : 4.68/14 : 16.03/16 in moles  

6.01 : 7.08 : .334 : 1.00

6.01/ .334  : 7.08/.334 : .334/.334 : 1.00/.334   in moles

18 : 21.19 : 1 : 3

18 : 21 : 1 : 3

empirical formula = C₁₈H₂₁NO₃

It is codeine .

7 0
3 years ago
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