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lubasha [3.4K]
2 years ago
14

How many moles of oxygen are needed to burn 425g of sulfur

Chemistry
2 answers:
Anika [276]2 years ago
3 0
I believe 212.5m, but I may be wrong, I’m a little rusty with moles
nataly862011 [7]2 years ago
3 0
I got 212.5m as well
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If 133mL of a 1.2 M glucose solution is diluted to 41.0L, what is the molarity of the diluted solution?
WITCHER [35]

the formula we is as follows:-

M1V1= M2V2

where

M1=1.2

V1=0.133l

V2=41l

M2=?

1.2 × 0.133 = 41 × M2

0.1596 = 41 × M2

M2 = 0.15960/41

M2 = 0.0038926829

7 0
3 years ago
Forces between similar molecules are said to be
podryga [215]

Answer:

Forces between similar molecules are said to be <em>cohesive</em> while those between different types of molecules are said to be <em>adhesive</em>.

Water 'beads' due to its strong <em>cohesive</em> forces. The meniscus of water in a glass tube is <em>concave</em> because the <em>adhesive</em> forces are strong.

Explanation:

The water in a tube has stronger adhesive forces between the water and glass molecules, so the cohesive forces between water molecules are weaker. That makes the water 'ascend' through the tube, giving a concave form of the meniscus. Another example is mercury, which is the opposite. In this case, the cohesive forces are stronger than the adhesive ones, thus the meniscus is convex.

4 0
3 years ago
Whats is not a characteristic of the eye of a hurricane
Marina86 [1]

B. has the storms strongest winds is not a characteristic of the eye of a hurricane because in the eye it is actually completely calm.

Hope that helps u!!

5 0
3 years ago
Read 2 more answers
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
2 years ago
Henry Moseley organized the periodic table by A. atomic weight. B. atomic number. C. electronegativity. D. overall charge of ato
Leni [432]
B - Atomic number. Dmitri Mendeleev organised the table according to atomic weight, however this caused problems with elements such as iodine and tellurium, Iodine has a higher mass, but a lower atomic number. And to make iodine in the same group as similar elements (halogens), Mendeleev had to break his own rules and put it before tellurium in the table. Moseley fixed this problem by ordering the elements according to atomic (proton) number.
3 0
3 years ago
Read 2 more answers
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