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Gelneren [198K]
3 years ago
6

What should a consumer consider when shopping for a credit card? Check all that apply.

Chemistry
2 answers:
Svetlanka [38]3 years ago
7 0
The initial fees to open the account.
The fees charged for late payment.
The annual percentage rate for the card.
Marina CMI [18]3 years ago
4 0

Answer:

B. the initial fees to open the account

C. the fees charged for late payments

E. the annual percentage rate for the card

Explanation:

Edg

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How many liters of 0.37 M solution can be made with 29.53 grams of lithium fluoride.<br> (LiF)?
dsp73

Answer:

V = 3.1 L      

Explanation:

Given data:

Molarity of solution = 0.37 M

Mass of LiF = 29.53 g

Volume of solution = ?

Solution:

Number of moles of LiF:

Number of moles = mass/molar mass

Number of moles = 29.53 g/ 25.94g/mol

Number of moles = 1.14 mol

Volume:

Molarity = number of moles of solute / Volume in L

0.37 M = 1.14 mol / V

V = 1.14 mol / 0.37 M

V = 3.1 L         (M = mol/L)

4 0
3 years ago
Why are relative atomic masses on the periodic tables not whole numbers?
Kobotan [32]

Answer:

They are averages.

Explanation:

atomic numbers on periodic tables are derived from the average value of all the isotopes of the element. So being averages they are sometimes not integers.

6 0
2 years ago
Read 2 more answers
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
3 years ago
Which one of the following solutions would have a pH value greater than 7?
Sliva [168]

Answer:

The 2 one

Explanation:

7 0
2 years ago
Based on the "Heating Curve for Wax," what is the boiling point for wax?
alukav5142 [94]
Looks like 340°C to me
4 0
2 years ago
Read 2 more answers
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