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2 years ago

What does multivalent mean???

Chemistry

1 answer:

salantis [7]2 years ago

4 0

In the context of multivalent ions, it is when it has multiple oxidative states.

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Draw a lewis structure for hnc and assign the non-zero formal charges to each atom. draw the lewis structures with the formal ch

Yuliya22 [10]

Answer: H-C-N H-N-C C-H-N Notice that C-H-N is the same as N-H-C just written backwards. ( i.e. they have the same connectivtiy.) You can exclude the last one with H in the middle since H has two bonds and 4 electrons around it. At this point you couldn't differentiate between the first two, so I would give you the connectivity in such a problem, which in this case is H-C-N.

3 0

3 years ago

How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*

MrMuchimi

Answer: 6.57 L of solution can be made.

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$ .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

$3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L$

Hence, 6.57 L of solution can be made.

5 0

3 years ago

Due is 10 min please help

Nonamiya [84]

Answer:

its c hope this helped;)

Explanation:

5 0

2 years ago

A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g

Lynna [10]

Answer:

$\%NaHCO_3=61.2\%$

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

$NaHCO_3+HA\rightarrow NaA+H_2O+CO_2$

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

$m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g$

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

$\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%$

Best regards.

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3 years ago

Someone please answer this in like 40 minutes Before it’s due. Explain the answer too

olga2289 [7]

Answer:

kkkwkisa

Explanation:

wulaoaiq8qiIq8qiaik

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2 years ago