2) If 100.0 J of heat are added to 20.0 g of water at 30.0°C, what will be the final

Which would be the best to neutralize a large acid spill in your school lab: sodium hydroxide or baking soda? Explain.

nadya68 [22]

Consider the acid spill. It is already starting to do nasty things to, say, the floor or counter. So you grab the bottle of 10% NaOH and pour some on the spill. All of a sudden, you get a great deal of heat, and you don't have any visual evidence whether your put on too little or too much. But you have added more liquid to the spill, generated more heat, and will get more damage. You have made a bigger mess, and if you added too much, you then have a neutralization problem to deal with.

And if it is something like a strong sulfuric acid solution, adding sodium hydroxide solution will be extremely exothermic, and you could get some really nasty results.

So now approach the spill with a handful of baking soda. You sprinkle it on the spill. It fizzes, and carbon dioxide is given off. That actually, in a very tiny way, moderates the temperature of the neutralization. And you can keep adding baking soda until the fizzing stops, and then perhaps some water to mix everything well. But what you have done is kept the volume to a minimum, added a neutralization agent that has a visible endpoint (no more gas being given off), and you don't suddenly have a huge amount of highly basic solution because you added too much.

And what is also nice about baking soda is that you can toss some with your hand or even with a spoon, and get some distance from the spill. With a liquid, you have to get much closer

i hope this helped..

5 0

3 years ago

Given the reaction: A + B <--> C + D

Lady_Fox [76]

Answer:

A.) 4.0

Explanation:

The general equilibrium expression looks like this:

$K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }$

In this expression,

-----> K = equilibrium constant

-----> uppercase letters = molarity

-----> lowercase letters = balanced equation coefficients

In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.

$K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }$ <----- Equilibrium expression

$K = \frac{[2 M] [2 M]}{[1 M] [1 M] }$ <----- Insert molarities

$K = \frac{4}{1 }$ <----- Multiply

$K = 4$ <----- Divide

6 0

2 years ago

Which element has the same number of valence electrons as calcium?

kipiarov [429]

Answer:

Barium has the same number of valence electrons as calcium

Explanation:

Valence electrons is the number of electrons of an atom on the outer shell.

Those valence electrons can participate in the formation of a chemical bond (if the outer shell is not closed); in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.

Calcium is an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons.

Sulfur is part of a group 16, called the chalcogens or oxygen family. Those atoms have 6 valence electrons. They can form a bound with atoms of group 2 such as calcium, but do not have the same number of valence electrons.

Potassium is part of group 1, called the alkali metals or lithium family. Those atoms have 1 valence electrons. That means Potassium do not have the same number of valence electrons like calcium.

Neon is part of group 18, the noble gasses. Those are stable atoms, which means they have 8 valence electrons. They do not have the same number of valence electrons like Calcium.

Barium an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons. Calcium is also part of this group.

This means barium has the same number of valence electrons as Calcium.

8 0

3 years ago

This is for a study guide, I can't figure it out!

miskamm [114]

The answer is "D" where the number of collisions per unit area is reduced by one-half. Drawing back on the piston means the volume is increased. The pressure is reduced. There are fewer collisions when the pressure is reduced.

3 0

3 years ago

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:

$0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY$

I hope it helps!

6 0

3 years ago