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MaRussiya [10]
3 years ago
8

A gold atom in its neutral state has how many electrons

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0
To find the number of electrons in a neutral gold atom look at it's atomic number. The atomic number is the number of protons in the element. For an atom to be neutral it needs the same number of electrons as protons, therefore the positive and negative charges will be balanced out. Gold's atomic number is 79 therefore it has 79 electrons.
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Match each of the following topical dosage forms with its correct definition.
Dmitry [639]

Answer:

cream - contains a higher proportion of oil than water

ointment - dr4g mixed in approximately equal proportions of oil and water

i don't know about the other two sorry

8 0
1 year ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
2 years ago
A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24
Setler79 [48]

<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

  • <u>For Part A:</u>

We are given:

p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg

To calculate the partial pressure of helium, we use the formula:

P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}

Putting values in above equation, we get:

745=245+119+163+p_{He}\\p_{He}=218mmHg

Hence, the partial pressure of Helium is 218 mmHg.

  • <u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g

Hence, the mass of helium gas is 0.504 g.

6 0
3 years ago
Given the equation representing a system at equilibrium in a sealed, rigid container:
LenaWriter [7]

Answer:

Choice 1. "HI to increase".

Explanation:

I found out the hard way.

4 0
3 years ago
Read 2 more answers
A nuclear fission reaction has mass difference between the products and the reactants of 0.0284 amu. Calculate the
notsponge [240]

Answer:

C) 4.24 x 10^{-12}

Explanation:

E = Δmc^{2}

Δm = 0.0284 x 1.66 x 10^{-27} kg = 4.714x10^{-29}kg

putting value in above equation

E = 4.714x10^{-29}kg x (3x10^{8})^{2} = 4.24 x 10^{-12}

6 0
3 years ago
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