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erik [133]
3 years ago
7

How many grams are in a 3.1E21 atom sample of vanadium

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0
There are 50.9415 grams in vanadium


hope that helps :)
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What is the most stable monatomic ion formed from magnesium?
Alekssandra [29.7K]

Answer:

Mg²⁺

Explanation:

The electron configuration of Mg is [Ne]3s².

It is easier for Mg to get a complete octet by losing two electrons and exposing the neon core than it is to add eight electrons and get an Ar octet.

The most stable monatomic Mg ion is Mg²⁺.

7 0
3 years ago
What is a substance?
ANTONII [103]
A is the answer im pretty sure
5 0
3 years ago
The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.
avanturin [10]

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

6 0
3 years ago
Suppose a system of two particles, represented by circles, have the possibility of occupying energy states with 0, 10, or 20 J.
MrMuchimi

Answer:

Explanation:

The possible energy states for the particles are 0, 10 and 20 J.  

The constraint in the system is that the total energy of the particles must be 20 J.

One given configuration where the total energy is 20 J is if both the particles occupy the 10 J state.

Hence, (10;10) is the given configuration.

Another possibility is if one of the particle is in 0 J state and another is in 20 J state. Hence, the system has a total energy of 0+20 = 20 J.

Hence, the possible configuration can be written as (0;20) or (20;0) which are energetically equivalent to the given configuration. Note that if the circles are indistinguishable, then the configuration (0,20) and (20,0) is the same thing.

4 0
3 years ago
How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL o
Rama09 [41]

Answer:

We need 75 mL of 0.1 M NaOH ( Option C)

Explanation:

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

Molarity of NaOH solution = 0.100 M

volume of 0.150 M CH3COOH = 50.00 mL = 0.05 L

<u>Step 2:</u> The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

<u>Step 3:</u> Calculate moles of CH3COOH

Moles CH3COOH = Molarity * volume

Moles CH3COOH = 0.150 M * 0.05 L

Moles CH3COOH =  0.0075 moles

<u>Step 4</u>: Calculate moles of NaOH

For 1 mol of CH3COOH we need 1 mol of NaOH

For 0.0075 mol CH3COOH we need 0.0075 mole of NaOH

<u>Step 5:</u> Calculate volume of NaOH

volume = moles / molarity

volume = 0.0075 moles / 0.100 M

Volume = 0.075 L = 75 mL

We need 75 mL of 0.1 M NaOH

7 0
3 years ago
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