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Murrr4er [49]
4 years ago
8

A pound is approximately 0.45 kilogram. A person weighs 87 kilograms. What is the persons weight, in pounds, when expressed to t

he correct number of significant figures.
Chemistry
1 answer:
pishuonlain [190]4 years ago
7 0
To work it out you’d go

87 ÷ 0.45
= 193.333… lbs

i don’t know how many significant figures you have to do but if it’s 3 significant figures then the answer is 193lbs

lbs = pounds
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The mass of a proton is 1.00728 amu andthat of a neutron is 1.00867 amu. What is the binding energy pernucleon (in J) of a Co nu
andreyandreev [35.5K]

Answer:

The binding energy per nucleon = 1.368*10^-12  (option D)

Explanation:

<u>Step 1:</u> Data given

The mass of a proton is 1.00728 amu

The mass of a neutron is 1.00867 amu

The mass of a cobalt-60 nucleus is59.9338 amu

Step 2: Calculate binding energy

The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.

Cobalt60 has 27 protons and 33 neutrons.

The mass of 27 protons = 27*1.00728 u = 27.19656 u

The mass of 33 neutrons = 33*1.00867 u = 33.28611 u

Total mass of protons + neutrons = 27.19656 u + 33.28611 u = 60.48267 u

Mass of a cobalt60 nucleus = 59.9338 amu

Mass defect = Δm = 0.54887 u

ΔE =c²*Δm

ΔE = (3.00 *10^8 m/s)² *(0.54887 amu))*(1.00 g/ 6.02 *10^23 amu)*(1kg/1000g)

Step 3: Calculate binding energy per nucleon

ΔE = 8.21 * 10^-11 J

8.21* 10^-11 J / 59.9338 = 1.368 *10^-12

The binding energy per nucleon = 1.368*10^-12  (option D)

3 0
3 years ago
Which organelle would be best represented by a refrigerator because it stores food and water?
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6 0
3 years ago
The equilibrium constant k for the synthesis of ammonia is 6.8x105 at 298 k. what will k be for the reaction at 375 k?
spayn [35]
Answer is: K <span>be for the reaction at 375 K is 326.
</span>Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
</span><span>Δ<span>H = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K</span></span></span>·mol.<span>
K</span>₁ = 6,8·10⁵.<span>
K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.
6 0
4 years ago
Read 2 more answers
Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi
Westkost [7]
1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
 

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2

Second, convert the number of moles to liters of gas at STP:

3.75 mol * 22.4 L/mol =  84 liters of H2

Answer: 84 liters

 </span>



6 0
4 years ago
Read 2 more answers
A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC
inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K

Putting values in above equation, we get:

\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0
3 years ago
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