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3 years ago

What are the sign and magnitude of the charge of atom containing 14 protons, 15 neutrons, and 12 electrons?

1 answer:

7 0

We know that protons gives positive charge, neutrons no charge and electrons negative, then +14 - 12 = +2 So the charge of the atom is sign positive and magnitude 2

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Consider the reaction 2Al(OH)3(s)→Al2O3(s)+3H2O(l) with enthalpy of reaction ΔHrxn∘=21.0kJ/mol What is the enthalpy of formation

bekas [8.4K]

Google knows your answer

8 0

3 years ago

Read 2 more answers

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t

Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0

3 years ago

If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn

pishuonlain [190]

Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.

[mW (m) - mW (g)]/mW (m) x 100% 

(they want % error so, if it is negative, just get rid of the sign)

3 0

3 years ago

JOSE BOUGHT 750 BAGS OF PEANUTS FOR 375 HE INTENDS TO SELL EACH BAG FOR 0.15 MORE THAN HE PAID HOW MUCH SHOULD HE CHARGE FOR EAC

labwork [276]

He paid 2 dollars for each bag. Then add 15 cents for that and he would charge $2.15 for each bag.

7 0

4 years ago

Read 2 more answers

How many grams of Co3+ are present in 2.34 grams of cobalt(III) nitrite?

Assoli18 [71]

Answer:

$m_{Co^{3+}}=0.563gCo^{3+}$

Explanation:

Hello there!

In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

$m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}$

Best regards!

4 0

3 years ago