how many moles of ammonium chloride are required for a complete reaction with 420 mol of calcium hydroxide

Many repeating, simple subunits are joined together. What is the most likely

Hatshy [7]

Answer:

O a polymer

Explanation:

When many repeating simple subunits are joined together, this results into a polymer.

The simplest unit or smallest unit of any substance is called a monomer. When many units of a monomer joins together, a polymer results.

For proteins, the monomeric unit is amino-acid.
When they combine they form longer chain molecules called proteins.
For carbohydrates, the monomeric unit is called glucose.
When they are combined they give us a wide range of carbohydrate molecules.

6 0

2 years ago

Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

7 0

2 years ago

How many moles of silicon are in 5.9 X 10 to the 24 power atoms of silicon

Dimas [21]

Answer:

9.79740949850 moles

Explanation:

1 mole = Avogardo's Number <<6.022 E 23 <<particles, atoms, etc.>>
This problem can be solved using dimensional analysis by multiplying atoms (5.9E24 atoms) by (1) mole and then dividing the number by Avogardo's number (6.022 E 23 atoms).
Note: E = * 10

Side Note: Please let me know if you need any clarifications about this!

8 0

2 years ago

Read 2 more answers

Which of the following equations has correct formulas and also is a balanced equation?

aliya0001 [1]

Answer : The correct option is, (B)

Explanation :

Balance equation : Balanced equation are those equation in which the number of atoms of each element and the total charge must be same on products side and reactants side.

In the given options, only option (B) equation has correct formula and also a balanced equation.

$2Be+O_2\rightarrow 2BeO$

while the other options has incorrect formulas and also an unbalanced equations.

The correct equation for option A is, $CaCO_3\rightarrow CaO+CO_2$

The correct equation for option C is, $2Mg+O_2\rightarrow 2MgO$

The correct equation for option D is, $KCl+HNO_3\rightarrow HCl+KNO_3$

Hence, the correct answer is, (B) $2Be+O_2\rightarrow 2BeO$

5 0

2 years ago

An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound

Temka [501]

The amount per 100 g is:

38.7 % calcium = 38.7g Ca / 100g compound = 38.7g

19.9 % phosphorus = 19.9g P / 100g compound = 19.9g

41.2 % oxygen = 41.2g O / 100g compound = 41.2g

The molar amounts of calcium, phosphorus and oxygen in 100g sample are calculated by dividing each element’s mass by its molar mass:

Ca = 38.7/40.078 = 0.96

P = 19.9/30.97 = 0.64

O = 41.2/15.99 = 2.57

C0efficients for the tentative empirical formula are derived by dividing each molar amount by the lesser value that is 0.64 and in this case, after that multiply wih 2.

Ca = 0.96 / 0.64 = 1.5=1.5 x 2 = 3

P = 0.64 / 0.64 = 1 = 1x2= 2

O = 2.57 / 0.64 = 4= 4x2= 8

Since, the resulting ratio is calcium 3, phosphorus 2 and oxygen 8

<span>So, the empirical formula of the compound is Ca</span>₃(PO₄)₂

5 0

2 years ago