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Flauer [41]
3 years ago
15

Can anyone please help

Chemistry
1 answer:
Naddik [55]3 years ago
7 0
All of them are reactants
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8.03 Solutions Lab Report<br> Does anyone have a PDF or Document of FLVS 8.03 Solutions Lab
GarryVolchara [31]

8.03 solutions report is described below.

Explanation:

8.03 Solutions Lab Report

In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.  

Pre-lab Questions:

In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.  

Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).

Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.

7 0
3 years ago
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Calculate the grams of C6H6 needed to produce 25 g of CO2<br> Show your work
Anna007 [38]

Answer:

The given reaction is a combustion reaction of benzene,

C

6

H

6

. From its balanced chemical equation,

2

C

6

H

6

+

15

O

2

→

12

C

O

2

+

6

H

2

O

,

the mass of carbon dioxide

(

C

O

2

)

produced from 20 grams (g) of

C

6

H

6

is determined through the molar mass of the two compounds, given by,

M

M

C

O

2

=

44.01

g

/

m

o

l

M

M

C

6

H

6

=

78.11

g

/

m

o

l

and their mole ratio:

12

m

o

l

C

O

2

2

m

o

l

C

6

H

6

→

6

m

o

l

C

O

2

1

m

o

l

C

6

H

6

With this,

m

a

s

s

o

f

C

O

2

=

(

20

g

C

6

H

6

)

(

1

m

o

l

C

6

H

6

78.11

g

C

6

H

6

)

(

6

m

o

l

C

O

2

1

m

o

l

C

6

H

6

)

(

44.01

g

C

O

2

1

m

o

l

C

O

2

)

=

(

20

)

(

6

)

(

44.01

)

g

C

O

2

78.11

=

5281.2

g

C

O

2

78.11

m

a

s

s

o

f

C

O

2

=

67.6

g

C

O

2

Therefore, the mass in grams of

C

O

2

formed from 20 grams of

C

6

H

6

is

67.6

g

C

O

2

.

it is a problem of app

3 0
2 years ago
How many valence electrons are in an atom of phosphorus
dezoksy [38]
5 Valence electrons .......... Hope it helps, Have a nice day:)
5 0
3 years ago
Read 2 more answers
What is the most striking part of this simulation?
oee [108]
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.

3 0
3 years ago
The frequency of an x-ray wave is 3.0 x 1012MHz. Its wave speed is 3.0x 108m/s. Calculate the wavelength of the x-ray wave below
jasenka [17]

Answer: The wavelength of the x-ray wave is 10^{-10}m

Explanation:

To calculate the wavelength of light, we use the equation:

\lambda=\frac{c}{\nu}

where,

\lambda = wavelength of the light  = ?

c = speed of x-ray= 3.0\times 10^8m/s

\nu = frequency of x-ray = 3.0\times 10^{12}MHz=3.0\times 10^{18}Hz= 3.0\times 10^{18}s^{-1}     (1Hz=1s^{-1})

Putting in the values:

\lambda=\frac{3.0\times 10^8m/s}{3.0\times 10^{18}s^{-1}}=10^{-10}m

Thus the wavelength of the x-ray wave is 10^{-10}m

5 0
3 years ago
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