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Gnom [1K]
2 years ago
6

Predict how maritime air masses would change if the ocean froze

Chemistry
1 answer:
MrRissso [65]2 years ago
3 0

Air moves from

high(pressure)the regions at the

poles(d0)

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13 This word describes the circulatory system that provides nutrients and oxygen to living cells of the body.
Nina [5.8K]

Answer: Cardiovascular System

Explanation:

This involves your heart, blood, veins, and arteries

4 0
3 years ago
The amount of a chemical solution is measured to be 2 liters.
irinina [24]
2 liters may be 1.5 to 1.9 rounded up to 2 or 2.1 or 2.4 rounded down to 2.

2 - 1.5 = 0.5

percent error = (absolute error / quantity) * 100

percent error = 0.5/2 * 100% = 0.25 * 100% = 25%

Choice C. 25%.
3 0
2 years ago
Ethanol (c2h5oh) melts at -114°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9
Nimfa-mama [501]

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid C_2H_5OH=-135^oC=138 K(0^oC=273K)

Melting temperature of Solid C_2H_5OH=114^oC=159 K

Temperature of liquid C_2H_5OH=-50^oC=223K

Specific heats of solid  ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid C_2H_5OH at 159 K

\Delta T_1 = 159 K - 138 K = 21 K

Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J

Heat required to melt and raise the temperature of C_2H_5OH upto 223 K

\Delta T_2 = 223 K - 159 K  = 64 K

Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J

Total heat to convert solid ethanol to liquid ethanol at given temperature :

Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

6 0
3 years ago
The acceleration of a runner who accelerates from 0 m/s to 3 m/s in 12s
Naily [24]
4 m/s^2


I hope this helps! 
5 0
3 years ago
15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what
s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a<em> limiting reactant problem.  </em>

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1</em>. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:        52.00   80.91       291.71

                2Cr  +  6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g:  15.00    15.00  

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

  Moles of Cr = 15.00 × 1/52.00

  Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×  

<em>Step 3</em>. Identify the<em> limiting reactant</em>  

Calculate the moles of CrCl₃ we can obtain from each reactant.  

<em>From Cr</em>:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

<em>From HBr: </em>

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

<em>Step 4</em>. Calculate the <em>theoretical yields</em> of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

   Theoretical yield of H₂ = 0.1854 × 3/6

   Theoretical yield of H₂ = 0.092 70 mol H₂

<em>Step 5</em>. Calculate the <em>volume of H₂</em> at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

<em>Step 6</em>. Calculate the <em>percent yield </em>

       % Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

       % yield = 15.35/18.03 × 100

       % yield = <em>85.15 % </em>

8 0
3 years ago
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