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Volume of Water -

miss Akunina [59]

Answer:

3.8 g/mL

Explanation:

From the question given above, the following data were obtained:

Volume of Water = 60 mL

Volume of Water + Object = 73.5 mL

Mass of object = 51.3 g

Density of object =?

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of Water = 60 mL

Volume of Water + Object = 73.5 mL

Volume of object =?

Volume of object = (Volume of Water + Object) – (Volume of Water)

Volume of object = 73.5 – 60

Volume of object = 13.5 mL

Finally, we shall determine the density of the object as illustrated below:

Volume of object = 13.5 mL

Mass of object = 51.3 g

Density of object =?

Density = mass /volume

Density of object = 51.3 /13.5

Density of object = 3.8 g/mL

Thus, the density of the object is 3.8 g/mL

5 0

3 years ago

2. Which number is not a coefficient in the equation,

prohojiy [21]

...........I guess D. 19

8 0

3 years ago

Read 2 more answers

Who said atoms are neither created nor destroyed in chemical reactions?

san4es73 [151]

John Dalton from 1803

8 0

3 years ago

A bus traveled south west for 422 kilometers. The entire trip took 3 hours. Calculate the velocity.

ahrayia [7]

V= d/t
v=speed
d=distance
t=time

So v= 422/3
v =140.667m/s

3 0

3 years ago

A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

seropon [69]

Answer: The p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, $M_1=(2\times 0.0031)=0.0062M$

Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, $M_2=(2\times 0.0042)=0.0084M$

To calculate the concentration of nitrate ions in the solution, we use the equation:

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

Calculating the p-function of zinc ions and nitrate ions in the solution:

For zinc ions:

$\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]$

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

For nitrate ions:

$\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]$

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

Hence, the p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

5 0

3 years ago

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