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statuscvo [17]
3 years ago
11

EXPLAIN: How does the amount of a substance affect the rate at which temperature changes?

Chemistry
1 answer:
Anton [14]3 years ago
4 0
The more of a substance you have the longer the reaction will take. If you have a big pot of water and a little pot of water. The little pot will heat up quicker then the bigger pot because it does not need to fill a space up. Think of it as diffusion. When you have a 500 and 50 mL beaker, and you add a blue and red dye to them both. It will only take about a minute for the dye to almost completely mix together in the 50 mL. While the 500 mL it will take much longer for dye to mix all together then it did in the 50 mL. When it is a bigger substance it takes longer for a reaction to fully expose the substance to that reaction. If you have food, if you only have one piece of food you only need to put it in for 30 seconds to 1 minute. While if you have multiple pieces of food you will need longer time to be able to heat all of the pieces. If you have a small bowl of soup and you put it in for a minute the middle will be warm. But if you had a bigger bowl and still put it in for only a minute then the middle will be cold because you did not give it enough reaction time to heat the whole substance.
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) Calculate the number of moles of sulfuric acid that is contained in 250 mL of 0.500 M sulfuric acid solution
maksim [4K]

Explanation:

According to the analysis, Molarity is amount mole per volume(1L). the amount in mole would be molarity × volume in litres.

0.500M × (250/1000)L= 0.125moles.

I hope this helps**

3 0
3 years ago
When you rub your palms together,
Roman55 [17]

Answer:

sliding friction occurs

4 0
3 years ago
Read 2 more answers
A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica
Sladkaya [172]

Answer: The empirical formula for the given compound is CH_5N

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = \frac{3.23}{3.23}=1

For Hydrogen  = \frac{16.2}{3.23}=5.01\approx 5

For Oxygen  = \frac{3.24}{3.23}=1.00\approx 1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is C_1H_5N_1=CH_5N

3 0
3 years ago
At what point is a base no longer added during a titration?
nordsb [41]

Answer:

when the pH increases suddenly in the flask

Explanation:

I think it's right. :)

6 0
2 years ago
Read 2 more answers
An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g
vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{1000 ml}{1 L}

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

4 0
3 years ago
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