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3 years ago

HELP PLZ!!!!!

1 answer:

6 0

The answer would be c as the cart is not in motion therefor ruling out kinetic and it is completely at rest making all of it energy potential

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What is the ph of a 0.01 m solution of the strong acid hno3 in water?

kumpel [21]

Answer is: pH of a 0,01 M solution is 2.
c(HNO₃) = 0,01 M = 0,01 mol/L.
pH = -log(c(HNO₃).
pH = -log(0,01 mol/L).
pH = 2.
pH is a numeric scale used to specify the acidity or basicity of an aqueous solution. If pH is less than seven, than solution is acidic and if pH is greater seven, solution is basic, if pH is equal seven, solution is neutral.

6 0

3 years ago

Read 2 more answers

How many molecules are in 21.5G of water

NemiM [27]

Answer:

There are 6.022×1023 molecules in a mole. There are 18.01528 grams of water per mole of water. in 1g water

Explanation:

please Mark my answer in brainlist

6 0

2 years ago

When an ionic compound dissolves in water, the individual ions first leave the crystal lattice and then each ion becomes surroun

vichka [17]

The entropy of the process in which the individual ions first leave the crystal lattice is positive while the entropy of the process whereby the each ion becomes surrounded by a cluster of polar water molecules is negative.

<h3>What is entropy?</h3>

The term entropy has to do with the degree of disorder in a system. The higher the entropy of the system, the more the disorderliness of the system.

Now, the entropy of the process in which the individual ions first leave the crystal lattice is positive while the entropy of the process whereby the each ion becomes surrounded by a cluster of polar water molecules is negative.

Learn more about entropy: brainly.com/question/13146879

3 0

2 years ago

Suppose you are helping mccarthy choose her sample sites. you have the resources to conduct the study at only 4 sites. pick the

VARVARA [1.3K]

I dont know the answer

4 0

3 years ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is $CO3^{-2}$ , and the ion nitrate is $NO3^{-}$ .

Sodium is in group 1, so it must lose one electron to be stable, and be the cation $Na^{+}$ . Silver has only one electron too, so the cation will be $Ag^{+}$ .

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0

3 years ago