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Vaselesa [24]
3 years ago
14

What were the seven different colors from newtons experiment

Chemistry
1 answer:
Ann [662]3 years ago
7 0
Red, Orange, Yellow, Green, Blue, Violet and Indigo
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18. Which one of the following is generally NOT true about elements in the
Dennis_Churaev [7]

Answer:

Statement B "Elements at the right side of the period are smaller in size" is not true.

Explanation:

Going through the rows of the elements in the period the size of the elements get continuously bigger.

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3 years ago
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Name the following compound:
krok68 [10]

Answer:

2-hexanone

Explanation:

First, we'll begin by:

1. Locating the longest continuous chain i.e haxane

2. Determine the functional group in the compound. The functional group in ketone (C=O). This changes the name from hexane to hexanone i.e replacing the -e in at the end in hexane with -one to make it hexanone.

3. Give the functional group the lowest low count. In doing this, we'll start counting from the left. The functional group is at carbon 2.

Note: no substitute group is attached.

Now, in naming the compound, we must indicate the position of functional group as illustrated below:

The name of the compound is:

2-hexanone

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4 years ago
Plaseee ASAP
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luh danm hahahahahahhaahah

7 0
3 years ago
What determines the velocity of an object
maxonik [38]

Answer:

the direction of the object

Explanation:

4 0
3 years ago
Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M aceto
GarryVolchara [31]

Answer:

(a) m=2.69m

(b) x_{LiBr}=0.099

(c) \% LiBr=18.9\%

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g

Next, we compute the mass of the solution:

m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g

Then, the mass of the solvent (acetonitrile) in kg:

m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg

Finally, the molality:

m=\frac{1.80mol}{0.670kg} \\\\m=2.69m

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol

Then, the mole fraction of lithium bromide:

x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099

(c) Finally, the mass percentage with the previously computed masses:

\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%

Regards.

4 0
4 years ago
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