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viktelen [127]
3 years ago
15

Calculate the final temperature when 50.0 mL of water at 65.0 °C are added to 25 mL of water at 25.0 °C.

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
7 0

Answer:

x=51.66

Explanation:

1 ml = 1 gram

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

Substituting values into the above, we then have:

(25)(25-x)(4.184)=(50)(x-65)(4.18)

Solve for x

x= 51.6667

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Which group of elements readily loses two electrons to form a compound?
nexus9112 [7]
The alkaline earth metals (the second group) because their ion charge is +2
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3 years ago
If you start with 89.3 g no(g) and 28.6 g h2(g), find the theoretical yield of ammonia.
Tatiana [17]
Balanced equation: 
<span>2 NO + 5 H2 ------> 2 NH3 + 2 H2O
 </span>
<span>2 moles NO react with 5 moles H2 to produce 2 moles NH3
 </span>
<span>Molar mass of NO = 30.00 g/mol </span>
<span>86.3g NO = 86.3/30.00 = 2.877 moles of NO </span>

<span>This will require: 2.877*5 / 2 = 7.192 moles of H2 </span>

<span>Molar mass of H2 = 2 g/mol </span>
<span>25.6g H2 = 25.6/2 = 12.7 mol H2. </span>
<span>You have excess H2 means the NO is limiting </span>

<span>From the balanced equation: </span>
<span>2 moles of NO will produce 2 moles of NH3 </span>
<span>2.877 moles of NO will produce 2.877 moles of NH3 </span>

<span>Molar mass NH3 = 17g/mol </span>
<span>Mass NH3 produced = 2.877 * 17 = 48.91g 

Hence the yield is = 48.91 g ~ 49 g</span>
3 0
3 years ago
Read 2 more answers
A 250.0-ml sample of ammonia, nh3 (g), exerts a pressure of 833 torr at 42.4 °c. what mass of ammonia is in the container
Lelu [443]
To solve this, let's assume ideal gas behavior.

PV=nRT
Let's solve for n. Convert units to SI units first.

Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K

n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol

The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>


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3 years ago
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Murljashka [212]

Answer 78

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3 years ago
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Natasha_Volkova [10]

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