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Fed [463]
2 years ago
11

What is the charge of the nickel ion in Ni(CIO3)3 if chlorate, ClO3-, has a charge of 1- ?*

Chemistry
1 answer:
o-na [289]2 years ago
5 0

Answer:

i am new but I haven't reached this level yet

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What types of damage can wind cause?
Levart [38]

Answer:

Wind between 39-46 mph can cause branches and limbs to break, make it hard for cars to stay on the road wind between 47-54 can cause lighting difficulties, wind between the 60-78 can cause trees to uproot and damage.

Explanation:

3 0
2 years ago
Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0
3 years ago
8y6× 9​4​​What number should Boris replace \blueD{y}ystart color #11accd, y, end color #11accd with?
Aleks [24]
Your question isn’t typed right :(
3 0
2 years ago
What does this graph represent?<br> A. speed<br> B. velocity<br> C. acceleration<br> D. momentum?
Llana [10]
I think that it might represent Velocity
5 0
2 years ago
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
Eva8 [605]

Answer:

m_{Ag}=135.8gAg

Y=88.4\%

Explanation:

Hello there!

In this case, according to the described chemical reaction, it is possible to compute the theoretical mass of silver as mass via the 1:2 mole ratio of copper to silver and their atomic mass in the periodic table, in order to perform the following stoichiometric setup:

m_{Ag}=40.gCu*\frac{1molCu}{63.55gCu}*\frac{2molAg}{1molCu}*\frac{107.87gAg}{1molAg}\\\\   m_{Ag}=135.8gAg

Next, given the actual yield of 120 g, we compute the percent yield via:

Y=\frac{120g}{135.8g}*100\%\\\\Y=88.4\%

Regards!

4 0
2 years ago
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