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3 years ago

Name at least two benefits of using models in science

Chemistry

2 answers:

scoray [572]3 years ago

8 0

Answer:

Viewing systems from multiple perspectives.

Discovering causes and effects using model tractability.Explanation:

good luck:)

STatiana [176]3 years ago

3 0

Answer:

Viewing systems from multiple perspectives.

Discovering causes and effects using model tractability.

Improving system understanding through visual analysis.

Explanation:

Got this from google, lol. But, I put three here just in case you could get extra credit for more than two.

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100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

2 years ago

What does 3.5 degree of unsaturation tell you?

Kaylis [27]

Answer:

It means the chemical entity is a radical

Explanation:

When we talk of unsaturation, we are referring to the number of pi-bonds in a chemical entity. The alkane, alkene and alkyne organic family are used to as common examples to explain the term unsaturation.

While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.

In the case of an alkene, there is only one single pi bond and as such the degree of unsaturation is 1.

Now in this case, we have a fractional 0.5 degree of unsaturation alongside the 3 to make a total of 3.5. So what’s the issue here?

The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.

Kindly recall that a radical is a chemical entity within which we have at the least an unpaired electron.

5 0

3 years ago

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$

$[A]=[A]_{0}e^{-kt}$

$ln[A]=ln[A]_{0} -kt$

$k=\frac{ln[A]_{0}-ln[A]}{t}$

therefore k is

$k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}$

8 0

3 years ago

For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re

vova2212 [387]

Answer:

A balanced ionic equation shows the reacting ions in a chemical reaction. These equations can be used to represent what happens in precipitation reactions or displacement reactions.

Precipitation reactions

In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

Ag+(aq) + Cl-(aq) → AgCl(s)

In a balanced ionic equation:

the number of positive and negative charges is the same

the numbers of atoms of each element on the left and right are the same

Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Question

Explain why this ionic equation is balanced:

Ba2+(aq) + SO42-(aq) → BaSO4(s)

Hide answer

There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

Question

Balance this ionic equation, which represents the formation of a silver carbonate precipitate:

Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Hide answer

2Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Question

Balance this ionic equation, which represents the displacement of iodine from iodide ions by chlorine:

Cl2(aq) + I-(aq) → I2(aq) + Cl-(aq)

Hide answer

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq

Explanation:

this will help, I used this for my work x

4 0

2 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago