Is a glow stick Endothermic or exothermic ?

What are 2 diseases caused by coronaviruses?

ivann1987 [24]

Answer:

SARS and MERS

Explanation:

8 0

2 years ago

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Hydrogen gas at a pressure of 740. mmHg has a volume of 2.00 L at a temperature of 25.0°C. What is the temperature of this gas a

nikdorinn [45]

Answer:

The final temperature of hydrogen gas is 537.63 K.

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 740 mmHg (740/760 = 0.97 atm)

Initial temperature = 25 °C (25 +273 = 298 K)

Final temperature =?

Final volume = 3.50 L

Final pressure = standard = 1 atm

Formula:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂T₁ / P₁V₁

T₂ = 1 atm × 3.5 L × 298 K / 0.97 atm × 2.00 L

T₂ = 1043 atm .L. K / 1.94 atm. L

T₂ = 537.63 K

7 0

2 years ago

A sample of 211 g of iron (III) bromide is reacted with

Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$

6 0

2 years ago

An acid is any substance that

ehidna [41]

Answer:

C

Explanation: a is incorrect since the lower the ph = more acidic and b is incorrect because it produces hydronium ion and d I’m not sure what it is but I no that base recieve the protons

4 0

2 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

2 years ago