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vodomira [7]
3 years ago
10

What volume of 0.436 M barium nitrate solution is needed to prepare 240.0 mL of a barium nitrate solution that is 0.300 M in nit

rate ion?
a. 349 mL

b. 82.6 mL

c. 55.0 mL

d. 165 mL

e. 330. mL
Chemistry
1 answer:
Advocard [28]3 years ago
7 0
The correct answer would be A

Explanation: I did the math
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Need anwser pls. Hurry!!!!! Please hurry up.<br>Hurry.​
wariber [46]
I’m sorry I have to do this
8 0
2 years ago
You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t
solniwko [45]

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

6 0
2 years ago
The teacher said the volume of the liquid was 500 mL when measured a student found it was 499.7 mL what is the students percent
Natalka [10]

Answer:

<h2>0.06 % </h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

error = 500 - 499.7 = 0.3

actual volume = 500 mL

We have

p(\%) =  \frac{0.3}{500}  \times 100 \\  =  \frac{3}{50}  \\

We have the final answer as

<h3>0.06 % </h3>

Hope this helps you

4 0
3 years ago
What geological processes form natural gas
Ivan
Decomposed organic matter ,mud,silt ,and sand 
5 0
3 years ago
Calculate the cfse for a d^3 system in an octahedral field in units of ∆_o. In other words, do not enter "∆_o" with your answer.
Rudik [331]

Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q

To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.

[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.

The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.

To learn more about crystal field stabilization energy  visit:brainly.com/question/29389010

#SPJ4

8 0
9 months ago
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