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77julia77 [94]
3 years ago
5

Bananas Foster is an example of a dessert that is flambéed. A Bananas Foster label states the accepted number of Calories to be

only 300 calories, but a calorimetry experiment measured there to be 318 calories. Calculate the percent error.
Chemistry
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

6.00%

Explanation:

Step 1: Given data

Accepted value for the number of calories in a Bananas Foster: 300 calories

Measured value for the number of calories in a Bananas Foster: 318 calories

Step 2: Calculate the percent error in the measure

We will use the following expression.

%error = |accepted value - experimental value|/ accepted value × 100%

%error = |300 cal - 318 cal|/ 300 cal × 100% = 6.00%

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A stock bottle of an acid reads: 56% by mass and 1.25 specific gravity. If the molar mass of the acid is 70g, find the mole per
TiliK225 [7]

Answer:

  • <u>10 mol / dm³</u>

Explanation:

Cubic decimeter is the same unit as liter; so, mole per cubic decimeter is mole per liter, and that is the unit of concentration of molarity. Thus, what is asked is the molarity of the solution. This is how you find it.

1. <u>Take a basis</u>: 1 dm³ = 1 liter = 1,000 ml

2. <u>Calculate the mass of 1 lite</u>r (1,000 ml) of solution:

  • density = mass / volume ⇒ mass = density × volume

Here, the density is given through the specific gravity

Scpecific gravity = density of acid / density of water

Take density of water as 1.00 g/ml.

  • density of solution = 1.25 g/ml

  • mass solution = 1.25 g/ml × 1,000 ml = 1,250 g

3. <u>Calculate the mass of solute</u> (pure acid)

  • % m/m = (mass of solute / mass of solution) × 100

  • 56 = mass of solute / 1,250 g × 100

  • mass of solute = 56 × 1,250g / 100 = 700 g

4. <u>Calculate the number of moles of solute</u>:

  • moles = mass in grams / molar mass = 700 g / 70 g/mol = 10 mol

5. <u>Calculate molarity (mol / dm³)</u>

  • M = number of moles of solute / liter of solution = 10 mol / 1 liter = 10 mol/liter.
5 0
3 years ago
A student mixed 115 g of sugar, 350 g of water and 5 g of spices. What will be the mass of the solution?
jenyasd209 [6]

Answer:

m_{solution}=470g

Explanation:

Hello,

In this case, a solution is formed when a solute is completely dissolved in a solvent, thus, for this situation, the sugar is the solute and the water the solvent but in addition to them we find spices which are also considered in the total mass of the solution. In such a way, for computing the total mass we must add the mass of three constituents (115 g sugar, 350 g water and 5 g spices) as shown below:

m_{solution}=115g+350g+5g\\\\m_{solution}=470g

Best regards.

5 0
3 years ago
If 100mL was diluted of a 2.0 M solution of sodium chloride by adding an additional 100 mL of water, what would the new molarity
andrew-mc [135]
I need help with my questions sorry
7 0
3 years ago
WILL GIVE BRAINLIEST
dedylja [7]

Answer:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c=4.18Jg∘C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.

In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.

What if you wanted to increase the temperature of 1 g of water by 2∘C ?

This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.

And there you have it. The equation that describes all this will thus be

q=m⋅c⋅ΔT , where

q - heat absorbed

m - the mass of the sample

c - the specific heat of the substance

ΔT - the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C

q=10,450 J

4 0
3 years ago
Which of the following causes a chemical change?
ycow [4]
Distinguish the difference between physical change or chemical change.

Should be C.
6 0
3 years ago
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