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Mamont248 [21]
2 years ago
5

# of protons = 48 Cadmium Titanium Oxygen Krypton

Chemistry
1 answer:
Dafna11 [192]2 years ago
7 0

Answer:

Cadmium +p= 48

Titanium p=22

Oxygen p=8

Krypton p=36

So the answer is Cadmium which has #48 protons

Explanation:

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Answer

                    Explanation:O2 = 1.7×10-3M;

N2 = 6.4×10-3M;

NO = 1.1 10-5M.

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Which taxonomic level includes the broadest characteristics?
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Your answer is kingdom
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Order the various instruments used for measuring water in this experiment (balances, graduated cylinders, beaker and pipette) fr
lara [203]

Answer:

(Most accurate) pippete>graduated cylinder>beaker>balance (Least accurate)

Explanation:

  1. <em>Most accurate. A pipette prived the most accurate method for delivering a known volume of solution, for example, a 10mL transfer pipette has an accuracy of ±0.02mL</em>
  2. A graduated cylinder is specifically used to deliver a known volume, its typical accuracy is ±1%, this means that a 100ml graduated cylinder is accurate to ±1mL.
  3. A beaker is a multipurpose cylindrical glass mainly used to hold liquids. Even though they are graduated, these marks are an estimation, the beaker's accuracy is around 10%.
  4. Least accurate. A balance measures an object's mass, even though water's density is close to 1, a balance is not the ideal equipment to measure volume, its capacity usually goes between 100-200grams and can measure mass to the nearest ±0.01mg to ±1mg.

I hope you find this information useful and interesting! Good luck!

4 0
3 years ago
For a solar eclipse to occur which of the following alignments is necessary? A. The moon is located along a straight line betwee
Aloiza [94]

For a solar eclipse t occur;

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8 0
3 years ago
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
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