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Ilia_Sergeevich [38]
3 years ago
13

Hey! need help with chem, multiple choice questions.

Chemistry
1 answer:
ladessa [460]3 years ago
7 0
1. 6 carbon atoms and triple bonding between carbons 2 and 3
2. i think it’s 3 but i may be wrong
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Which describes the slope of the graph from 10 s to 30 s?
inna [77]

Answer:

3

Explanation:

Use the rise over run method. plot the necessary points first in the x and y axis

6 0
3 years ago
Anyone tryna help mee
devlian [24]

Answer:

It is a solid

4 0
3 years ago
A compound that is composed of molybdenum (Mo) and oxygen (O) was produced in a lab by heating molybdenum over a Bunsen burner.
irakobra [83]
First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
     Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo

We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
     Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g 

We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
     Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O

To convert mass to moles, we use the molar mass of each element.
     1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
     0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O

0.0131 mol is the smallest number of moles. We divide each mole value by this number:
     0.0131 mol Mo / 0.0131 = 1
     0.0200 mol O / 0.0131 = 1.53

Multiplying these results by 2 to get the lowest whole number ratio,
     0.0131 mol Mo / 0.0131 = 1 * 2 = 2
     0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.
5 0
3 years ago
Read 2 more answers
Ice
fomenos
I think I it’s gonna be C.
8 0
3 years ago
PLEASE HELP
earnstyle [38]
Answer:
               Cp  =  0.093 J.g⁻¹.°C⁻¹
Solution:

The equation used for this problem is as follow,

                                                  Q  =  m Cp ΔT   ----- (1)

Where;
            Q  =  Heat  =  300 J

            m  =  mass  =  267 g

            Cp  =  Specific Heat Capacity  =  ??

            ΔT  =  Change in Temperature  =  12 °C

Solving eq. 1 for Cp,

                                 Cp  =  Q / m ΔT


Putting values,
                                 Cp  =  300 J / (267 g × 12 °C)

                                 Cp  =  0.093 J.g⁻¹.°C⁻¹
6 0
3 years ago
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