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3 years ago

Why is the finance charge important to consider when shopping for credit?

1 answer:

5 0

The correct option is C.
Finance charge on credit card refers to the interest you pay for borrowing money when you are using a credit card. It is the interest you are charged on the debts you owe and sometimes it might include other charges such as penalty fees for late payment. Individual finance charge is usually calculated using the the person APR [Annual Percentage Rate], amount of debt owe and the time period been considered. The higher your debts, the higher your finance charge.

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What is the mass in grams of 2.6 moles of angelic acid?

Dennis_Churaev [7]

Answer:

260.34g

Explanation:

First, you need to know what angelic acid is comprised of. It is written as C₅H₈O₂.

In order to solve for the mass of 2.6 moles of angelic acid, you need the mass of 1 mole of angelic acid. This can be found by adding the masses from the periodic table, like shown below:

5 carbon atoms = (5)(12.01g) = 60.05g

8 hydrogen atoms = (8)(1.01) = 8.08g

2 oxygen atoms = (2)(16) = 32g

angelic acid = 60.05 + 8.08 + 32 = 100.13g

Then, set up a basic stoichiometric equation and solve. The units should cancel out.

$(\frac{2.6mol}{1} )(\frac{100.13g}{1mol} )=260.34g$

5 0

3 years ago

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.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

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:

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

For process 2 :

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

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

If 120 g of potassium chloride are mixed with 100 g of water at 80°C, how much will not dissolve?

djverab [1.8K]

Answer:

yes

Explanation:

How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-.

4 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L