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SVEN [57.7K]
2 years ago
14

12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

Chemistry
1 answer:
Kaylis [27]2 years ago
3 0

Answer:

115.625^{\circ}\text{F}

Explanation:

m_1 = First mass of water = 12 oz

m_2 = Second mass of water = 20 oz

\Delta T_1 = Temperature difference of the solution with respect to the first mass of water = (T-75)^{\circ}\text{F}

\Delta T_2 = Temperature difference of the solution with respect to the second mass of water = (T-75)^{\circ}\text{F}

c = Specific heat of water

As heat gain and loss in the system is equal we have

m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}

The final temperature of the solution is 115.625^{\circ}\text{F}.

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N2 + 3 H2 → 2 NH3
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B

Explanation:

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How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?
NARA [144]

Answer:

18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

Explanation:

Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.

The decomposition of mercury oxide is given by the chemical equation below:

2HgO ----> 2Hg + O₂

2 moles of HgO decomposes to produce 1 mole of Hg

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18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg

Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

3 0
2 years ago
How do you solve this problem
ad-work [718]

Answer: Volume of gas in the stomach, V = 0.0318L or 31.8mL

Explanation:

The number of moles of oxygen will remain constant even though the liquid oxygen will undergo a change of state to gaseous inside the person's stomach due to an increase in temperature.

<em>Number of moles of oxygen gas = mass/molar mass</em>

molar mass of oxygen gas = 32 g/mol

mass of oxygen gas = density * volume

mass of oxygen gas = 1.149 g/ml * 0.035 ml

mass of oxygen gas = 0.040215 g

Number of moles of oxygen gas = 0.0402 g/(32 g/mol)

Number of moles of oxygen gas = 0.00125 moles

<em>Using the ideal gas equation, PV=nRT</em>

where P = 1.0 atm, V = ?, n = 0.00125 moles, R = 0.082 L*atm/K*mol, T = (37 + 273)K = 310 K

<em>V = nRT/P</em>

V = (0.00125moles) * (0.082 L*atm/K*mol) * (310 K) / 1 atm

V = 0.0318L or 31.8mL

3 0
3 years ago
A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21 c and the pressure is lowered to 0.80 atm,
Nuetrik [128]

Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

\frac{P*V}{T}=k

Having an initial state 1 and a final state 2 it is possible to say that:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

  • P1= 1 atm
  • V1= 1.2 L
  • T1= 273 °K
  • P2= 0.80 atm
  • V2= ?
  • T2= 21°C= 294 °K

Replacing:

\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}

Solving:

V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}

V2= 1.62 L

<u><em>The new volume is 1.62 L</em></u>

<u><em></em></u>

8 0
2 years ago
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