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Sever21 [200]
3 years ago
6

The radioisotope that has the longest half-life is the best to use in powering planet and space exploration vehicles because the

y can travel farther. Which radioisotope is the best to use?

Chemistry
2 answers:
adoni [48]3 years ago
8 0

Answer:

D. plutonium-239

Explanation:

BaLLatris [955]3 years ago
4 0

Answer : The radioisotope which has longest half-life and can be best to be used in powering planet and space exploration vehicle and travel farther will be Plutonium - 239 as it has the half life of 24065 years. Which in comparison to other radioisotopes is longest.


Hence, Plutonium - 239 can be the best suitable radioisotope for powering the planet and space exploration vehicle to travel farther.

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When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

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

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

7 0
3 years ago
How much heat must your body transfer to 500.0g of water to heat it from 25.0°C to body temperature, 37.0°C?
shtirl [24]

Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of water = 500.0 g

c = specific heat of water = 4.18J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 37.0^oC

Now put all the given value in the above formula, we get:

Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K

Q=25080J=25.1kJ

Therefore, the heat your body transfer must be, 25.1 kJ

3 0
3 years ago
Please help man i’ll reward brainliest ✊
zheka24 [161]

Answer:

gAIN

Explanation:

6 0
3 years ago
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 m
IrinaK [193]

Answer:

49.45~%

Explanation:

In this case, we have to start with the <u>chemical reaction</u>:

C_6H_1_2O~->~C_6H_1_0~+~H_2O

So, if we start with <u>10 mol of cyclohexanol</u> (C_6H_1_2O) we will obtain 10 mol of cyclohexanol (C_6H_1_0). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

(6*12)+(10*1)=82~g/mol

With this value we can calculate the grams:

10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0

Now, we have as a product 500 mL of C_6H_1_0. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

500~mL\frac{0.811~g}{1~mL}=405.5~g

Finally, with these values we can calculate the <u>yield</u>:

%~=~\frac{405.5}{820}x100~=~49.45%%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

6 0
3 years ago
Which best describes the primary function of the nervous system in keeping the body in homeostasis
mojhsa [17]

HEY BUDDY THE ANSWER IS

It processes the environment and sends out signals.

4 0
4 years ago
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