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2 years ago

12

Which surface has the least amount of friction?

2 answers:

Bogdan [553]2 years ago

8 0

Which surface has the least amount of friction?

An ice rink

A grassy field

A paved road

A sidewalk

Answer:

<h2>An Ice rink</h2><h2 />

hram777 [196]2 years ago

4 0

Answer:

An ice rink has the least amount of friction.

Any smooth surface has the least amount of friction.

I hope this helped at all.

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Which electrons participate in chemical bonding?

OverLord2011 [107]

B.) Valence Electrons. The nucleus of the valence electrons attracts and pulls atoms together.

7 0

2 years ago

Read 2 more answers

Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w

quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0

2 years ago

Liquid does not act that a solid. how is it different from a solid and gas?

Wittaler [7]

Unlike solid matter, where particles are tightly packed and slightly vibrating, or gas, where particles go around everywhere and are extremely loose, a liquid has particles that are loosely packed but are still in slight contact with each other. Hope that's good enough

6 0

2 years ago

MgCl2 + AgNO3 → AgCl + Mg(NO3)2

Law Incorporation [45]

Answer:

MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2

Explanation:

I'm assuming you want to balance it so...

The first thing I see is that there are two chlorines on the reactant side and one on the product side

Adding a coefficient of 2 would get 2AgCl2

Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.

If that is not what you are looking for let me know!

6 0

2 years ago

Read 2 more answers

How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?

Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

$pH= -log[H]\\pH= -log (\frac{kw}{[OH]})$

$8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}$

$[OH]=1.69E-6$

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

$Cu(OH)_2 -> Cu^{2+} + 2OH^-$

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

$Kps= [Cu^{2+}] [OH]^2$

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

$Kps= s*(2s+1.69E-6)^2$

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is $s=4.96E-8 mol/L$

Now, let us calculate the moles in 1 L:

$moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles$

7 0

3 years ago