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svet-max [94.6K]
3 years ago
6

How many moles are in 2.23 X 10^50 molecules of H2?

Chemistry
1 answer:
prisoha [69]3 years ago
7 0

Answer:2.23

⋅

10

50

X

,

H

2

Explanation:

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Winter is coming and it's time to make sure you have the right amount of antifreeze
ololo11 [35]

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of  ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

  • mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

  • mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

  • mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

  • molality

\tt =\dfrac{107.608}{6}=17.93~m

5 0
3 years ago
How is it that when a salt sample dissolves in water, the delta S for the process is positive?
Bess [88]

Answer:-

Water is highly ordered. In water each oxygen atom is connected to others around it through hydrogen bonding via bridging hydrogen atoms. When a salt like NaCl is dissolved, some of these Hydrogen bonds break.

When a salt like NaCl dissolves in water, the NaCl breaks in to ions Na+ and Cl-.

The water molecules now surround these ions.

The slightly negative oxygen end of water molecule gets near the Na+, while the slightly positive Hydrogen of water molecule gets near the Cl-.

So before salt sample dissolve, the water molecules were highly ordered due to hydrogen bonding. Now after salt dissolve there is a decrease in order and thus an increase in disorder of the water molecules.

Due to increase in disorder, entropy which is a measure of disorder increases. Since entropy increases, delta S for the process is positive.

4 0
3 years ago
In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch
LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of copper (II) chloride.

M_2\text{ and }V_2 are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?

Putting values in above equation, we get:

0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL

Hence, the volume of stock solution needed are, 12.5 mL

8 0
3 years ago
You have to decide to change the temperature at which you run a certain reaction in hope of obtaining more product more quickly.
Alexandra [31]

Answer:

see explanation

Explanation:

The reaction has a negative rate law; i.e., Rate = - ΔConcentration / ΔTime which is graphically a negative slope for the plot of Rate as a function of reactant concentration. => Rate ∝ f(Reactant Concentration). However, by raising the temperature, an increase the probability of reaction occurs and the formation of more product.

4 0
3 years ago
Draw a Lewis structure for the following molecules on scratch paper. Then give the number of bonds between the C and N in each.
Yuliya22 [10]

Refer to the attachments

6 0
3 years ago
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