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pshichka [43]
2 years ago
12

Need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow!!

Chemistry
1 answer:
marin [14]2 years ago
6 0

Answer:

Q14: 17,140 g = 17.14 kg.

Q16: 504 J.

Explanation:

<u><em>Q14:</em></u>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).

m is the mass of the ice (m = ??? g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).

∵ Q = m.c.ΔT

∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)

∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.

<u><em>Q16:</em></u>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 12.0 g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).

∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.

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djyliett [7]

Answer:

22.7%

Explanation:

Given that last year stock for company A was $7200

The stock for company B last year was worth $3510

Stock in company A decreased by 24%

This means the new value of stock for company A became;

(100-24)/100 *$7200

76/100*$7200

0.76*7200 =$5472

Stock in company B decreased by 20%

This means the new value of stock for company B became;

(100-20)/100 *$3510

80/100*$3510

0.8*$3510

$2808

Original investors stock value was = $7200+$3510 =$10710

New investors stock value is = $5472+$2808=$8280

Decrease in value of stock = $10710-$8280 =$2430

percentage decrease in stock value = decrease in stock/original value of stock *100%

=2430/10710 *100 =22.689

=22.7%

7 0
2 years ago
A gaseous mixture of 10.0 volumes of CO₂, 1.00 volume of unreacted O₂, and 50.0 volumes of unreacted N₂ leaves an engineat 4.0 a
ycow [4]

The masses of CO and CO2​ are 90.55g and 100−90.55=9.45 g respectively.

<h3>Total mass.</h3>

Let the mixture has 100g as total mass.

The number of moles of CO is 2890.55​=3.234.

The number of moles of CO2​ is 449.45​=0.215.

The mole fraction of CO is 3.234+0.2153.234​=0.938.

The mole fraction of CO2​ is 1−0.938=0.062.

The partial pressure of CO is the product of the mole fraction of CO and the total pressure.

It is 0.938×1=0.938 atm.

The partial pressure of carbon dioxide is 0.062×1=0.042 atm.

The expression for the equilibrium constant is:

Kp​=PCO2​​PCO2​​=0.062(0.938)2​=14.19

Δng​=2−1=1

Kc​=Kp​(RT)−Δn=14.19×(0.0821×1127)−1=0.153.

To learn more about equilibrium constant visit the link

brainly.com/question/15118952

#SPJ4

3 0
1 year ago
What pressure (in atm) will 0.44 moles of co2 exert in a 2.6 l container at 25°c?
Lilit [14]
We can use the ideal gas law equation to find the pressure 
PV = nRTwhere 
P - pressure 
V - volume  - 2.6 x 10⁻³ m³ 
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³  = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm 
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm
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3 years ago
16. The structure shown is representative of which substituted hydrocarbon?
Viktor [21]

Answer:

The answer is B. Ether

Explanation:

An atom of Oxygen between a carbon chain is called Ether

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The oxidation state of Mn2
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The oxidation state of Mn2 is 2+
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