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yaroslaw [1]
2 years ago
10

What portion of your salt solution in ice water system is: a. A mixture? b. A solution?

Chemistry
1 answer:
ahrayia [7]2 years ago
8 0

The salt and water are a homogeneous mixture but when salt dissolves in the water system is called a solution of salt and water.

<h3>What is a mixture? </h3>

A mixture is defined as the combination of two or more substances that are not chemically bonded together.

There are two types of mixture which include:

  • Homogeneous (uniform composition) and

  • Heterogeneous mixtures

When salt is added to the ice water system, it lowers the freezing point of the ice water thereby forming a homogenous mixture of water and salt.

The dissolution of salt in ice water leads to the formation of salt and water solution.

Learn more about mixture here:

brainly.com/question/10677519

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How will the price of a product impact on the profits of the entrepreneur​
taurus [48]

Answer:

The higher your price, the less volume you have to produce for a given dollar amount of profit! Even a small price increase can generate significant additional profit. ... When a business comes out with a new product or service and they are the first to market, they may be able to charge high prices initially.

Explanation:

3 0
3 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

7 0
3 years ago
How might proper lab techniques impact a scientist work?
lana [24]
Because if the the technique is wrong the scientist is wrong I’m sorry it’s a bad answer :(
7 0
3 years ago
Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear rea
vesna_86 [32]

Answer:

The correct answer is - 4.

Explanation:

As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -

The sum of the product mass number of products = mass of reactant

237Np93 →233 Pa91 +AZX is the equation,

Solution:

Mass of reactants = 237

Mass of products are - Pa =233 and A = ?

233 + A = 237

A = 237 - 233

A = 4

So the equation will be:

237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)

5 0
3 years ago
How is a suspension different from a colloid?
Mashcka [7]

Answer:

Explanation:

Colloidal:

  • Colloid consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm.  
  • The particles in colloid can not be seen through naked eye.
  • It is homogeneous.
  • We can not separate the colloidal through the filtration. The pore size of filter paper is 2μm. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.

Suspension:

  • The particle size in suspension is greater than 1000 nm.
  • These particles can seen through naked eye.
  • It is heterogeneous.  
  • The particles in suspension can be separated through the filtration.
3 0
3 years ago
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