A solid has an volume

Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10

Sophie [7]

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

(1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

$p_{A} =9.98g/10.00mL=0.998g/mL\\p_{B} =15.61g/15.00mL=1.041g/mL\\p_{C} =25.65g/25.00mL=1.026g/mL$

(2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

$p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3} \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL$

(3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error= $\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100$

%error= $\frac{|1.022g/mL-0.997655g/mL|}{0.997655g/mL}*100$

%error=2.44 %

7 0

3 years ago

A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated nitric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted nitric acid solution

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

Hence, the volume of concentrated solution required is 9.95 mL

6 0

2 years ago

What is the percent yield for the reaction of nitrogen and hydrogen to produce ammonia, if the theoretical yield was 12.5 g but

koban [17]

Answer:

81.6%

Explanation:

10.2/12.5= 0.816

0.816=81.6%

4 0

2 years ago

How are coefficients used to balance chemical equations?

Zepler [3.9K]

According to the law of conservation of mass, the quantity of the elements, involved in chemical reactions does not change. For example,
H2O2 - > H2O + O2
is wrong, because there are two O atoms on the first side of the equation, and three on the other. To correct it, coefficients must be added, until the amount of both H and O atoms is equal on both sides.
2H2O2 - > 2H2O + O2

4 0

2 years ago

What mass of solute must be used to prepare 500ml of 0.100M aqueous sodium borate Na2B4O7 from solid hydrated sodium borate Na2B

motikmotik

0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)

Explanation:

First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles

We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.

Now to find the mass of hydrated sodium borate we use the following formula:

number of moles = mass / molar weight

mass = number of moles × molar weight

mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g

Learn more about:

molar concentration

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4 0

2 years ago