Answer:
They are called homogenous mixtures. If you put sand into a glass of water, it is considered to be a mixture. You can always tell a mixture, because each of the substances can be separated from the group in different physical ways. You can always get the sand out of the water by filtering the water away.
Answer:
Empirical formula of C₈H₈ = CH
Explanation:
Data Given:
Molecular Formula = C₈H₈
Empirical Formula = ?
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule
C₈H₈ Consist of Carbon (C), and Hydrogen (H)
Now
Look at the ratio of these two atoms in the compound
C : H
8 : 8
Divide the ratio by two to get simplest ratio
C : H
8/8 : 8/8
1 : 1
So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1
So the empirical formula will be
Empirical formula of C₈H₈ = CH
The net ionic equation formed is
Ag^+(aq)+Cl^−(aq)→AgCl(s)
Chromium(III) nitrate and silver(I) chloride are the products of the balanced molecular equation for the reaction between chromium(III) chloride and silver(I) nitrate. An (s) next to the chemical formula for silver(I) chloride designates it as an insoluble salt.
CrCl3(aq)+3AgNO3(aq)→Cr(NO3)3(aq)+3AgCl(s)
Silver and the chloride ions are the two ions that must interact to create silver(I) chloride. By designating ions as the reactants and silver(I) chloride as the product, the net ionic equation is formed.
Ag^+(aq)+Cl^−(aq)→AgCl(s)
Ionic Equation:
In general, anions and cations react to generate a compound in a dissolved media, which is known as an ionic reaction. Water-insoluble salts are created when the ions of water-soluble salts interact with one another in an aqueous media.
To learn more about Ionic equaion click the given link
brainly.com/question/19705645
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Answer:
The correct answer is "Secondary active transport".
Explanation:
Secondary active transport is a form of across the membrane transport that involves a transporter protein catalyzing the movement of an ion down its electrochemical gradient to allow the movement of another molecule or ion uphill to its concentration/electrochemical gradient. In this example, the transporter protein (antiporter), move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. The 3 Na⁺ are the ions moved down its electrochemical gradient and the one Ca⁺⁺ is the ion moved uphill its electrochemical gradient, because Na+ and Ca⁺⁺are more concentrated in the solution than inside the cell. Therefore, this scenario is an example of secondary active transport.
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