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3 years ago

7

What is the mass percentage composition of O in L-carnitine (C7H15NO3), a compound that is taken as a dietary supplement to re-

duce muscle fatigue?
Answer in units of %. Your answer must be within ± 3.0%

1 answer:

sveta [45]3 years ago

4 0

Answer:

3.5% not sure but thank it is

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Which of the following characterizes an ionic bond?

velikii [3]

Answer:

option A

I think so good night sweet dreams

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3 years ago

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A 5.00-g sample of copper metal at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the copper is

vekshin1

<u>Answer:</u> The final temperature of the copper is 95°C.

<u>Explanation:</u>

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

$Q= m\times c\times \Delta T$

Q = heat absorbed = +133 J (heat is added to the system)

m = mass of copper = 5.00 g

c = specific heat capacity of copper = 0.38 J/g ° C

$\Delta T={\text{Change in temperature}}=T_2-T_1$

$T_1=25^oC$

Putting values in above equation, we get:

$+133J=5.00g\times 0.38J/g^oC\times (T_2-25)\\\\T_2=95^oC$

Hence, the final temperature of the copper is 95°C.

3 0

3 years ago

A battery can provide a current of 4.60 A at 3.40 V for 2.50 hr. How much energy (in kJ) is produced? 1st attempt kJ Energy

Lunna [17]

Answer:

The energy produced equals 140.760 kJ

Explanation:

The relation between power, current and voltage is

$Power=Current\times Voltage$

Applying the given values in the relation above we get

$Power=4.60\times 3.40=15.64W$

Now Since $Power=\frac{Energy}{Time}\\\\Energy=Power\times Time$

Again applying the calculated values we get

$Energy=15.64\times 2.50\times 3600=140760Joules=140.76kJ$

4 0

3 years ago

A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter

Ira Lisetskai [31]

Answer:

m H2O = 56 g

Explanation:

Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

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3 years ago

How many moles of nitrogen are contained in 18.65 L at STP?

stira [4]

1 mole of any gas under STP ----- 22.4 L

18.65 L*1 mol/22.4 L ≈ 0.8326 mol N2

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3 years ago

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