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3 years ago

10

What was the original pressure in a rigid container if the new pressure is 82.5atm and the temperature was raised from 44C to 67

C

1 answer:

zhenek [66]3 years ago

4 0

Answer: 54 atm

Explanation:

I did 67/82.5 then got 0.8121212121. I them divided 44 by 0.81212121 and got 54.1791044776

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How many grams of KOH do we need to make 250.0ml of a 0.75 M

ZanzabumX [31]

Answer:

10.52g KOH

Explanation:

250.0 ml X 1L/1000ml X 0.75 mol KOH/1L X 56.105gKOH/1 mol KOH =10.52g KOH

5 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp = 9.05×10−18?

saveliy_v [14]

<span>M(NO3)2 ==> [M2+] + 2 [NO3-] 0.202 M ==> 0.202 M M(OH)2 ==> [M2+] + 2[OH-] 5.05*10^-18 ===> s + [2s]^2 5.05*10^-18 ===> 0.202 + [2s]^2 5.05*10^-18 = 0.202 * 4s^2 4s^2 = 25*10^-18 s^2 = 6.25*10^-18 s = 2.5*10^-9 So, the solubility is 2.5*10^-9</span>

5 0

3 years ago

Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2

Serga [27]

Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

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I hope it helps!

5 0

3 years ago

What is the composition of Gilsonite?

nekit [7.7K]

Answer:

Roughly C100 H140 N3 O

Explanation:

Gilsonite is a bituminous product that resembles shiny black obsidian.

It contains more than 100 elements.

Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.

Its empirical formula is roughly C100 H140 N3 O.

3 0

3 years ago