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Aleksandr [31]
3 years ago
15

A hockey puck moves 22 meters northward, then 16 meters southward, and finally 4 meters northward.For this motion, what is the d

istance moved?What is the magnitude and direction of the displacement? *
Chemistry
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

Answer is

The distance moved is : 42m

The magnitude and direction of the displacement: 10m

Explanation:

The distance moved is:

22+16+4=42

The magnitude and direction of the displacement:

22-16+4=10

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

6 0
3 years ago
A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267
blsea [12.9K]
atomic mass=percentage of isotope a * mass of  isotope a + percentage of isotope b * mass of  isotope b+...+percentage of isotope n * mass of isotope n.

Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%

mass of isotope₂=270.9 u
percentage of isotope₂=9.7%

Therefore:

atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u

Answer: the mass atomic of this element would be 268.1 u
7 0
3 years ago
3. How can a community help implement Ecological Solid Waste Management.
Setler79 [48]

Answer:

It help cause we compost food scraps and other organic wastes. We also reuse and recycle materials to organize for government and industry to develop community recycling materials.

It depend on your opinion as a student.

Explanation:

Hope it helps

4 0
2 years ago
9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
2 years ago
2 icl + h2i2 + 2 hcl is first order in icl and second order in h2. Complete the rate law for this reaction in the box below. Use
OleMash [197]

2 ICl + H2   ----> I2 + 2 HCl

as given that rate is first order with respect to ICl and second order with respect to H2

The rate law will be

Rate = K [ICl] [ H2]^2

b) Given that K = 2.01 M^-2 s^-1

Concentrations are

[ICl] = 0.273 m and [H2] = 0.217 m

Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s

5 0
3 years ago
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