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3 years ago

12.2 scientific notation

Chemistry

1 answer:

Nadya [2.5K]3 years ago

3 0

This would be 1.22 x 10^1

You simply move the decimal.

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I answered a question correctly and someone deleted my answer without giving me any explanation. I don't know why someone would

lina2011 [118]

Answer:

Same, my answer got deleted. Its a troll on here who wants to screw around with people.

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3 years ago

Balance the equation below and use it to answer this question:

Svet_ta [14]

Answer:

2H2 + O2 -----> 2H2O

Not sure about the second question though.

8 0

3 years ago

Calculate the vapor pressure at 25 ?c of hexane above the solution. the vapor pressure of pure hexane at 25 ?c is 151 torr.

Studentka2010 [4]

One of the many ways in order to solve for the vapor pressure of pure components at a given temperature is through the Antoine's equation which is written below,
P = 10^(A - B/C+T)
where A, B, and C are constants and T is the temperature in °C and P is the vapor pressure in mm Hg.
For hexane,
A = 7.01
B = 1246.33
C = 232.988

Substituting the known values,
P = 10^(7.01 - 1246.33/232.988+25)
<em> P = 151.199 mm Hg</em>

8 0

3 years ago

Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperat

KonstantinChe [14]

Answer:

$\boxed{\text{139 $\, ^{\circ}$C}}$

Explanation:

The question is asking, "At what temperature does the vapour pressure of water equal 3.4 atm?"

To answer this question, we can use the Clausius-Clapeyron equation:

$\ln \left (\dfrac{p_{2}}{p_{1}} \right) = \dfrac{\Delta_{\text{vap}}H}{R} \left(\dfrac{1 }{ T_{1} } - \dfrac{1}{T_{2}} \right)$

Data:

p₁ = 1 atm; T₁ = 373.15C

p₂ = 3.4atm; T₂ = ?

R = 8.314 J·K⁻¹mol⁻¹

$\Delta_{\text{vap}}H = \text{39.67 kJ//mol}$

(The enthalpy of vaporization changes with temperature. Your value may differ from the one I chose.)

Calculation:

$\begin{array}{rcl}\ln \left (\dfrac{p_{2}}{p_{1}} \right)& = & \dfrac{\Delta_{\text{vap}}H}{R} \left( \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right)\\\\\ln \left (\dfrac{3.4}{1} \right)& = & \dfrac{39670}{8.314} \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\\ln3.4 & = & 4771 \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\1.224 & = & 12.78 - \dfrac{4771}{T_{2}}\\\\\dfrac{4771}{T_{2}} & = & 11.56\\\\\end{array}$

$T_{2} & = & \dfrac{4771}{11.56} = \text{412.6 K} = \textbf{139 $\, ^{\circ}$C}\\\\\text{The maximum temperature is } \boxed{\textbf{139 $\, ^{\circ}$C}}$

7 0

3 years ago

NH,F+ _AICI,<br>__NH,CI + __A!F,

mart [117]

Explanation:

NH₄F+ AICI₃, → NH₄CI + AlF₃

NH,F: in ionic form

NH₄⁺ + F⁻ = NH₄F

AICI in ionic form:

Al³⁺ + Cl⁻ = AlCl₃

NH Cl in ionic form:

NH₄⁺ + Cl⁻ = NH₄Cl

Al F in ionic form:

Al³⁺ + F⁻ = AlF₃

NH₄F+ AICI₃, → NH₄CI + AlF₃

Now we have to balance the equation with respect to the atoms:

aNH₄F+ bAICI₃, → cNH₄CI + dAlF₃

for N: a = c

H: 4a = 4c

F: a = 3d

Cl: 3b = c

let a = 1

c = 1

b = $\frac{1}{3}$

d = $\frac{1}{3}$

Therefore we can imply that:

a = 3, b = 1 , c = 3 and d = 1 if we multiply through by 3

3NH₄F+ AICI₃, → 3NH₄CI + AlF₃

Learn more:

Chemical equation brainly.com/question/2924195

#learnwithBrainly

8 0

3 years ago