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guapka [62]
3 years ago
13

How many atoms is 100.0g of aluminum?

Chemistry
1 answer:
34kurt3 years ago
8 0

Answer:

How many moles are present in 100.0 g of Al? What is the mass of 0.552 mol of Ag metal? Answer a: 3.706 mol

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Your brain and spinal cord are joined to the other parts of your body through __________.
Lostsunrise [7]

Answer is A

You never run down adjacent to you spinal cord and bones. That’s where they are the most protected. There are also holes in the pelvis bone for nerves to pass through

5 0
3 years ago
What is the percent by mass of oxygen in carbon dioxide (CO2)?
wariber [46]

Answer:

= 72.73%

Explanation:

The percentage by mass of an element is given by;

% element = total mass of element in compounds/molar mass of compound × 100

The mass of oxygen in carbon dioxide = 32 g

Molar mass of CO2 = 44 g

Therefore;

% of O2 = 32/44 × 100%

             <u>= 72.73%</u>

8 0
3 years ago
PlEASE HELP! 40!
podryga [215]

Answer:

moles Na = 0.1114 g / 22.9898 g/mol=0.004846

moles Tc = 0.4562g /98.9063 g/mol=0.004612

mass O = 0.8961 - ( 0.1114 + 0.4562)=03285 g

moles O = 0.3285 g/ 15.999 g/mol=0.02053

divide by the smallest

0.02053/ 0.004612 =4.45 => O

0.004846/ 0.004612 = 1.0 => Tc

to get whole numbers multiply by 2

Na2Tc2O 9

Explanation:

Hope it right hope it helps

5 0
2 years ago
How many grams of Cl are in 345 g of CaCl2
swat32
Hope this helps you.

3 0
3 years ago
Determine the ionic strength, μ, for each of the solutions. Assume complete dissociation of each salt and ignore any hydrolysis
kow [346]

Answer:

The answers are: (a )0.0058 M; (b) 0.00681 M; (c) 0.006912 M

Explanation:

Ionic strenght= μ= 1/2 ∑ c z²

Where c is the concentration of each ion (in M) and z is the charge of the ion. So, to calculate the ionic strenght of a solution you have to know the concentration of each ion and their charges. For this, the dissociation equilibrium is required.

a) HCl →    H⁺    +    Cl⁻

Conc H⁺= 0.00580 M

Conc Cl⁻= 0.00580 M

μ= 1/2 x ((Conc H⁺)(+1)²) + ((Conc Cl⁻)(-1)²

μ= 1/2 x ((0.00580 M) + (0.00580 M))

μ= 0.00580 M

b) CaBr₂  →  Ca²⁺ + 2 Br⁻

Conc Ca²⁺= 0.00227 M

Conc Br⁻= 2 x 0.00227 M= 0.00454 M

μ= 1/2 ((Conc Ca²⁺)(+2)² + ((Conc Br⁻)(-1)²)

μ= 1/2 (0.00227 M (4)) + (0.00454 M)

μ= 0.00681 M

c) The solution has two dissociation equilibria:

Mg(NO₃)₂ → Mg²⁺ + 2 NO₃⁻

La(NO₃)₃ → La³⁺ + 3 NO₃⁻

Conc Mg²⁺= 0.000752 M

Conc NO₃⁻= 2 x (0.000752 M) + 3 x (0.000776 M)= 0.003832 M

Conc  La³⁺= 0.000776 M

μ= 1/2 ((Conc Mg²⁺)(+2)²)+ ((Conc NO₃⁻)(-1)²) + ((Conc  La³⁺)(+3)²)

μ= 1/2 (0.000752 M(4)) + (0.003832 M) + (0.000776 M)(9))

μ= 0.006912 M

6 0
3 years ago
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