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Tomtit [17]
3 years ago
5

How much heat is required to raise the temperature of 40.0 g of aluminum from 50.0 degrees C to 90.0 degrees C? The specific hea

t of aluminum is 0.895 joules per gram C
Chemistry
1 answer:
Alex73 [517]3 years ago
4 0

Answer:

The answer to your question is  Q = 1432 Joules

Explanation:

Data

mass of Al = m = 40.0 g

temperature 1 = T1 = 50°C

temperature 2 = T2 = 90°C

Specific heat = C = 0.895 J/g°C

heat = Q = ?

Process

1.- Write the formula to calculate heat

              Q = mC(T2 - T1)

2.- Substitution

              Q = 40(0.895)(90 - 50)

3.- Simplification

              Q = 40(0.895)(40)

4.- Result

               Q = 1432 Joules

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Leni [432]

Answer:

Explanation:

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3 0
3 years ago
If you mixed three Dixie cups’ contents together containing 0.05L of 1.0M lemonade, 0.05L of 2.5M lemonade, and 0.05L of 0.5M le
viva [34]

Molarity of the resulting solution will be 1.33 M.

<u>Explanation:</u>

First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume

For cup 1  = 1  M ×0.05 L = 0.05 moles

For cup 2 = 2.5 M × 0.05 L= 0.125 moles

For cup 3 = 0.5 M × 0.05 L = 0.025 moles

Total moles = 0.05 + 0.125 + 0.025  = 0.2 moles

We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L

Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M

3 0
3 years ago
How many atoms in 7.55 mole of Iron atom
vagabundo [1.1K]

7.55 x 6.02 x 10²³ = 4.55 x 10²⁴ atoms

4 0
2 years ago
A sample of argon has a volume of 0.165L at -34.0°C and a 0.98atm of pressure. What would the volume of this gas at STP be?
JulijaS [17]

Answer: 0.185 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.98 atm

P_2 = final pressure of gas = 1 atm  (at STP)

V_1 = initial volume of gas = 0.165 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = -34.0^oC=273-34.0=239.0K

T_2 = final temperature of gas = 273K  (at STP)

Now put all the given values in the above equation, we get:

\frac{0.98\times 0.165}{239}=\frac{1\times V_2}{273}

V_2=0.185l

Thus the volume of this gas at STP be 0.185 L

8 0
3 years ago
What do the following have in common: MgCl2, AlF3, CaI2, KCl
arsen [322]

Answer:

They are salts, ionic compounds

Explanation:

To know if they are covalent or ionic we need to check their difference in electronegativity. As an example mgcl2, mg has electronegativity of 1.2 and cl 3 so 3-1.2=1.8 which is bigger than 1.7 means it's ionic therefore they are salts

8 0
3 years ago
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