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3 years ago

Jake put a bottle of water on the desk and left the room. When he came

Chemistry

1 answer:

fiasKO [112]3 years ago

4 0

Answer:

The air in the room

Explanation:

The diffrent temp of the water compared to the room. The water was colder than the room.

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Increasing the concentration of a reactant shifts the position of a chemical equilibrium towards formation of more products. wha

Natalija [7]

<h2>The required "option is e)".</h2>

Explanation:

The Rate of formation of products depends on the concentration of reactants or the forward reaction increases on increasing the rate of the concentration.
There is no effect on the equilibrium rate when the concentration of reactants and products is constant.
Forward reaction slows down when the reactant concentration is decreased.
On increasing the amount or concentration of reactant the chemical equilibrium shifts towards formation of more products.
Hence, the forward reaction speeds up.

4 0

3 years ago

i'm reasking this, i need help with this and it's due today. the answer isn't 30,240 minutes, it's the blanks i need filled in :

adelina 88 [10]

Answer:

Explanation:

24hours ×21 days=

you will get the answeer

3 0

3 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

3 years ago

A mixture of carbon dioxide and helium gases is maintained in a 7.91 L flask at a pressure of 1.42 atm and a temperature of 33 °

Crank

Answer:

The gas mixture contains 1.038 grams of helium

Explanation:

Step 1: Data given

Volume of the flask = 7.91 L

Total Pressure = 1.42 atm

Temperature = 33 °C

Mass of CO2 = 8.25 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of He = 4 g/mol

Step 2: Calculate total number moles of gas

p*V = n*R*T

⇒ p = the pressure = 1.42 atm

⇒ V = the volume = 7.91 L

⇒ n= the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L* atm/K*mol

⇒ T = the temperature = 33 °C = 306 Kelvin

n = (p*V)/(R*T)

n = (1.42*7.91)/(0.08206 * 306)

n = 0.447 moles

Step 3: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 8.25 grams / 44.01 g/mol

Moles CO2 = 0.1875 moles

Step 4: Calculate moles of Helium

Moles Helium = total moles of gas - moles of CO2

Moles Helium = 0.447 - 0.1875 = 0.2595 moles of helium

Step 5: Calculate mass of helium

Mass of helium = moles of helium * molar mass of helium

Mass of helium = 0.2595 moles * 4 g/mol

Mass of helium = 1.038 grams

The gas mixture contains 1.038 grams of helium

8 0

3 years ago

The mechanism for producing a concentrated urine involves all of the following except

avanturin [10]

Answer:

The correct answer is "obligatory water reabsorption in the proximal convoluted tubule".

Explanation:

The mechanism for producing concentrated urine cannot include the obligatory reabsorption of water in the proximal convoluted tubule since this process is part of the nephron, the system that filters the blood. Glucose and amino acids are reabsorbed almost entirely, as are approximately 70% of filtered potassium and 80% of bicarbonate.

Have a nice day!

4 0

3 years ago