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3 years ago

14

You are writing a safety contract for your class. List 10 things you would include in the contract?

1 answer:

maria [59]3 years ago

8 0

Answer:

1.always listen to teacher

2 no eating in class

3 always have attentive listening

4 always have proper safety material

5 wear eyeglasses when needed

6 let others be able to listen

sorry I could only think of 6

Explanation:

I just have six for you sorry...

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You are given two unknown acids, HX and H2Y, and two values for Ka, 1x10^-7 and 1x10^9. You know that X is a much more electrone

Elena L [17]

Answer:

.

Explanation

In HX , X is more electronegative than Y so HX will ionise more because of ionic bond between H and X . On the other hand H₂Y will be less polar as compared to HX so it will ionise to a lesser extent . Hence Ka will be more for HX . Ka represents the degree of ionisation of acid . Higher the ionisation , higher is the value of Ka . H₂Y which is less polar will ionise less and hence it will have lesser value of Ka .

Hence H₂Y will have value of 10⁻⁷ and HX will have value of ka equal to 10⁹ .

6 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

Questions for buffers lab, please help.

TEA [102]

Answer:

because the acid properties of aspirin may be problematic.

4 0

2 years ago

What does pressure measure

omeli [17]

Answer:

C

Explanation:

Pressure it says it in the name its the force of colliding particles

6 0

3 years ago

When 33 g of CaO and 10 g of H2O react, how many grams of calcium hydroxide would you expect to be produced* Explain your answer

MrMuchimi

Answer:

41.44 g

Explanation:

First of all, we must put down the equation of the reaction;

$CaO + H2O ----->Ca(OH)2$

Number of moles of CaO = 33g/56 g/mol = 0.59 moles

Number of moles of H20 = 10g/18 g/mol = 0.56 moles

Since the reaction is in 1:1 mole ration, H2O is the limiting reactant

Hence;

mass of Ca(OH)2 produced = 0.56 moles * 74 g/mol = 41.44 g

7 0

3 years ago

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