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Nana76 [90]
3 years ago
11

The factors in an experiment are called what

Chemistry
2 answers:
DanielleElmas [232]3 years ago
7 0
The factors in an experiment are called variable
koban [17]3 years ago
6 0

Answer:

Variables

Explanation:

A variable is any factor, trait, or condition that can exist in differing amounts or types.

Hope this helps :  )

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Suppose that the C=O group in a peptide bond can be regarded as isolated from the rest of the molecule. Given that the force con
Stella [2.4K]

Answer:

a) v = 1497.2 cm^-1

b) v = 1465 cm^-1

Explanation:

In the attached image is the procedure explained to reach the answer.

4 0
3 years ago
Draw the R enantiomer of CFClBrI as a 3-D (wedge and dotted line) structure and as a Fischer projection (putting the I at the to
vazorg [7]
I don’t know This one is there a picture ?
4 0
3 years ago
Solve the following equation (y = 1.2345x – 0.6789) for x, given that y = 0.570
Andrews [41]

x = 1.01

Explanation:

Given equation:

   y = 1.2345x – 0.6789

   y = 0.570

Problem:

Solving for x

The variables in this equation are y and x

They can take up any value since they are variables.

  Since we have been given y = 0.570

        y = 1.2345x – 0.6789

To solve for x, we simply substitute for y in the equation;

     since  y = 0.570

     0.57 = 1.2345x – 0.6789

    add 0.6789 to both sides;

 0.57 + 0.6789 =  1.2345x – 0.6789 + 0.6789

  1.2489‬ =   1.2345x

Divide both sides by  1.2345

 \frac{1.2489}{  1.2345} = \frac{  1.2345x }{ 1.2345}

    x = 1.01

learn more:

Equations brainly.com/question/9045597

#learnwithBrainly

 

7 0
3 years ago
The density of no2 in a 4.50 l tank at 760.0 torr and 25.0 °c is ________ g/l.
kap26 [50]
From  ideal  gas  equation   that   is   PV=nRT
n(number of  moles)=PV/RT
P=760 torr
V=4.50L
R(gas  constant =62.363667torr/l/mol
T=273 +273=298k
n  is   therefore   (760torr x4.50L) /62.36367 torr/L/mol  x298k  =0.184moles
the  molar  mass  of  NO2 is  46  therefore  density=  0.184  x  46=8.464g/l
7 0
3 years ago
Read 2 more answers
Be sure to answer all parts. calculate δg ocell for the reaction between cr(s) and cu2+(aq). e ocell = 1.08 j/c. enter your answ
Mademuasel [1]

Answer:

\boxed{-6.29 \times10^{5}\text{ J}}

Explanation:

Step 1. Determine the cell potential

                                                  <u>    E°/V     </u>

2×[Cr ⟶ Cr³⁺ + 3e⁻]                  0.744  V

<u>3×[Cu²⁺ + 2e⁻ ⟶ Cu]             </u>   <u>0.3419 V </u>

2Cr + 3Cu²⁺ ⟶ 3Cu  + 2Cr³⁺    1.086  V

Step 2. Calculate ΔG°

\Delta G^{\circ} = -nFE_{\text{cell}}^{^{\circ}} = -6 \times 96 485 \times 1.086 = \text{-629 000 J}\\\\= \boxed{-6.29 \times10^{5}\text{ J}}

6 0
3 years ago
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