With the given problem you gave here, I can't answer the question because I need more details. Luckily, I found a similar problem that's provided with a diagram and a table shown in the attached picture.
This test is called the Benedict's test which is used as test for presence of sugars. If the solution contains sugar, like glucose, the solution would turn from blue to red. If not, it would stay blue.
<em>Therefore, the correct results would be that in row 3.</em>
To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.
35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22
Average atomic mass closest to 35.22 amu.
Answer : The concentration of Si in kilograms is, 
Explanation :
As we are given that, the concentration of Si in an Fe-Si alloy is 0.25 wt% that means:
Weight of Si = 0.25 g = 0.00025 kg
Weight of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg
Density of Si = 
Density of Fe = 
Now we have to calculate the concentration in kilograms of Si per cubic meter of alloy.
Concentration of Si in kilograms = 
Concentration of Si in kilograms = 
Now put all the given values in this expression, we get:
Concentration of Si in kilograms = 
Concentration of Si in kilograms =
Thus, the concentration of Si in kilograms is,
Answer:
See explanation below
Explanation:
What we have to consider is the hybridation of the three carbon atoms we are asked in this question .
Hybridization # bonds Angle
sp³ 4 109.5º
sp² 3 + 1 pi bond 120º
sp 2 + 2 pi bonds 180º
Carbon atom (a) is bonded to two atoms: Carbon (b) and one Hydrogen. It has a triple bond to Carbon (b). Therefore its hybridization is sp with two pi bonds, and for sp hybridization we know the angle is 180 º.
The same hybridization sp happens to carbon (b) bonded to Carbon (c) and C(a) using one sp bond to Carbon (a) and 2 pi bonds; it is also bonded using the other sp to Carbon (c). The angle is therefore 180 between Carbons b and c.
Carbon C is bonded to 4 atoms, therefore, its hybridization is sp³ and the angles with these 4 atoms will be 109.5 º tehedral ( one bond to OH, one to C(b), and 2 to H ) .
Answer:
0.97 mole
Explanation:
1 mole will give 6.02×10^23 atoms
Xmole of tungsten will give 5.82×10^23 atom of tungsten
X= 5.82×10^23/ 6.02×10^23
X = 0.97 moles of tungsten
