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skad [1K]
3 years ago
6

What is Zeff of phosphorus

Chemistry
1 answer:
professor190 [17]3 years ago
5 0

Answer:

Explanation:

Here is an illustration showing how to "calculate" +5 as the effective nuclear charge (Zeff) for phosphorus. We are not implying that the electrons are in orbits here, this is simply showing inner shell electrons that shield the outer valence electrons from the full nuclear charge.

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Write formulas and names for two binary molecular compounds
lyudmila [28]

Answer:

NaCl sodium chloride

HF hydrogen flouride

Explanation:

7 0
3 years ago
What is the unknown metal if the temperature of a beaker of 100ml of water was raised 17c to 19 c when 21 grams of the metal at
horrorfan [7]

Answer:

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

Explanation:

<u>Step 1</u>: Data given

Mass of the metal = 21 grams

Volume of water = 100 mL

 ⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams

Initial temperature of metal = 122.5 °C

Initial temperature of water = 17°C

Final temperature of water and the metal = 19 °C

Heat capacity of water = 4.184 J/g°C

<u />

<u>Step 2: </u>Calculate the specific heat capacity

Heat lost by the metal = heat won by water

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)

21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)

-2173.5 *c(metal) = -836.8

c(metal) = 0.385 J/g°C

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

4 0
4 years ago
i’m in desperate need of help haha. btw it’s 200 ml of a clear liquid and 50 ml of a different clear liquid. please help
Lelu [443]
C. a chemical reaction occurred when the two liquids were mixed
6 0
3 years ago
If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

5 0
3 years ago
A catalyst decreases the activation energy of a particular exothermic reaction by 34 kJ/mol, to 57 kJ/mol. Assuming that the mec
vagabundo [1.1K]

Answer:

the activation energy for the uncatalyzed reverse reaction is 83kJ/mol

Explanation:

see the attached file

4 0
3 years ago
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