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Lady_Fox [76]
3 years ago
14

When planning a simple experiment, what does it mean to “test one variable”?

Chemistry
1 answer:
SSSSS [86.1K]3 years ago
3 0

Answer:

I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."

Explanation:

I remember from last year but I'm not totally sure. Good luck!

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Chemistry!! Please help!!
cluponka [151]

Answer:

4.993 ×10⁻¹¹ J

Explanation:

The <em>nuclear binding energy</em> is the energy equivalent to the mass defect.

The <em>mass defect</em> is the difference between the mass of a nucleus and the sum of the masses of its nucleons.

<em>Calculate the mass defect </em>

  16 p = 16 × 1.007 28 u =  16.116 48  u

  16 n = 16 × 1.008 67 u =  16.138 72  u

Total mass of nucleons = 32.255 20 u

             - Mass of S-32 = <u>31.972 070 u </u>

                 Mass defect =  0.283 13   u

Convert the <em>unified atomic mass units to kilograms</em>.

Mass defect

= \text{0.283 13 u} \times \frac{ \text{1.66 06} \times 10^{-27} \text{ kg}}{\text{1 u}}

= 4.700 \times 10^{-28} \text{ kg}

Use Einstein’s equation to <em>convert the mass defect into energy</em>

E = mc^{2}

E = 4.7000 \times 10^{-28} \text{ kg} \times (2.998 \times 10^{8} \text{ m}\cdot\text{s}^{-1})^{2} = 4.224 \times 10^{-11} \text{ J}

4 0
3 years ago
In which chemical system is molecule to ion attractions present? A) KNO3(s) B) KNO3(l) C) KNO3(aq) D) KNO3(g)
Maurinko [17]

Answer:

KNO_3 (aq)

Explanation:

Potassium nitrate is a soluble salt which readily dissolves in a polar solvent, such as water. When solid potassium nitrate is dissolved in water, it dissociates into potassium cations and nitrate anions.

Due to the resultant ionic charges, the polar water molecules attract the resultant ions and potassium nitrate ions become hydrated, that is, surrounded by water molecules.

Nitrate, our anion, attracts the partially positive ends of water molecules by attracting them via hydrogen atom.

Potassium, the cation, attracts the partially negative end of water molecules by attracting via oxygen atom.

3 0
3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
1.2 moles of (NH4)3PO3
Aleks04 [339]
1.2 moles of (nph4)3po3 is.......159.6 grams
3 0
3 years ago
As a human being, we are considered to be a ..... because we rely on other organism for food​
ryzh [129]
Heterotrophs i think :)

3 0
3 years ago
Read 2 more answers
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