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3 years ago

A rock is dropped into a graduated cylinder filled with 35 mL of water.

Chemistry

1 answer:

avanturin [10]3 years ago

7 0

Answer:

5 cm³.

Explanation:

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

Volume of water = 35 mL

Volume of water + Rock = 40 mL

Volume of Rock =.?

The volume of the rock can be obtained as follow:

Volume of Rock = (Volume of water + Rock) – (Volume of water)

Volume of Rock = 40 – 35

Volume of Rock = 5 mL.

Finally, we shall convert 5 mL to cm³.

This can be obtained as follow:

1 mL = 1 cm³

Therefore,

5 mL = 5 mL × 1 cm³ / 1 mL

5 mL = 5 cm³

Thus, we can say that the volume of the rock is 5 cm³.

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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.

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3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

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3 years ago

What are alkenes used to make?

evablogger [386]

Answer:

By far the most important use of alkenes is in the making of plastics as plastics are used in almost everything.

Explanation:

Alkenes themselves aren't used much in everyday life however Alkenes are very important to industrial synthesis as it is relatively easy to turn them into other things.

Alkenes can be turned into polymers or plastics through addition reactions and the most common ethene is turned into everything from plastic bags to bottles.

Alkenes can also be turned into alcohols. most commonly propene is used as a feedstock to produce butanol and other products useful in industry or for production

Explanation:

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3 years ago

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What is a supersaturated solution?

oee [108]

Answer:

A supersaturated solution is a more solute solution than can be dissolved by the solvent.

Explanation:

sodium acetate is an example of one

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2 years ago

Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction

Luden [163]

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

The chemical equation for this reaction is:

HA + B ⇌ A⁻ + HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

There are no Brønsted-Lowry acids and bases in this reaction.

Reason: In this reaction, there is no exchange of proton between the acid and the base.

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3 years ago