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3 years ago

HELP ME ASAP PLEASE LOTS OF POINTS

Chemistry

1 answer:

DerKrebs [107]3 years ago

4 0

Answer: 32.06 atomic mass units

Explanation:

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An analytical chemist is titrating 181.2 mL of a 0.09000M solution of diethylamine ((CH) NH) with a 0.5400M solution of HNO3. Th

gavmur [86]

Answer:

874 por que mi pen3 ne tu ocncha

Explanation:

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3 years ago

Can anyone please help me find molar masses of compounds. For example Copper (ll) sulfate (CuSO4)

Alina [70]

Answer:

Explanation:

You would have to add up the atomic masses of all the compounds in the compound, making sure you include how many molecules of each are in the compound

For example, in CuSOA we have 1 molecule of Cu and S, as 4 molecules of O

The atomic masses are as follows:

Cu = 63.55 u

S = 32.065 u

O = 15.99 units

This is how we would add it up:

(Atomic mass of Cu) + (Atomic mass of S) + 4(Atomic Mass of O)

(63.55) + (32.065) + 4(15.99)

(63.55) + (32.065) + 63.96

= 159.575 u

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2 years ago

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How many grams of oxygen are required to burn 0.10mole of c3h8?

Anastaziya [24]

1mol—44g/mol
0.10mol—x
x=0.10*44
x=4.4 g

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3 years ago

Microorganisms and humus have little impact on soil health. Please select the best answer from the choices provided T F

egoroff_w [7]

First of all what the heck to the other guy that answered. And the correct answer is FALSE :))

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3 years ago

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Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)

AURORKA [14]

Explanation:

The net equation will be as follows.

$K(s) + Cl_{2}(g) \rightarrow KCl(s)$

So, we are required to find $\Delta H_{formation}$ for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1: Convert K from solid state to gaseous state

$K(s) \rightarrow K(g)$ , $\Delta H_{1}$ = 89 kJ

Step 2: Ionization of gaseous K

$K(g) \rightarrow K^{+}(g) + e^{-}$ , $H_{2}$ = 418 KJ

Step 3: Dissociation of $Cl_{2}$ gas into chlorine atom .

$\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$ , $\Delta H_{3} = \frac{244}{2}$ = 122 KJ

Step 4: Iozination of chlorine atom.

$Cl(g) + e^{-} \rightarro Cl^{-}(g)$ , $H_{4}$ = -349 KJ

Step 5: Add $K^{+}$ ion and $Cl^{-}$ ion formed above to get KCl .

$K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s)$ , $H_{5}$ = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

$\Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}$

= 89 + 418 + 122 - 349 - 717

= - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

5 0

4 years ago