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How many molecules of water are there in 0.234 L of water

tankabanditka [31]

Answer:

7.82x10^24 molecules of water

Explanation:

H2O=18.015 g/mol Avogadro's #=6.022x10^23 molecules

0.234L x 1000g/1L x 1 mol H2O/18.015 g x 6.022x10^23 = 7.82x10^23 molecules of water

3 0

2 years ago

A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo

zheka24 [161]

Explanation:

The given data is as follows.

Width of Styrofoam = 24.0 cm

Length of Styrofoam = 36.0 cm

Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

Volume = length × width × height

= (36.0 × 24.0 × 5.0) $cm^{3}$

= 4320 $cm^{3}$

or, = $4.32 \times 10^{3} cm^{3}$

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

24.0 cm × 36.0 cm × 2.0 cm

= $1.73 \times 10^{3} cm^{3}$

Hence, mass of displaced water is as follows.

mass = density × volume

= $1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}$

= $1.73 \times 10^{3} g$

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

Mass of water displaced = mass of book + mass of Styrofoam

$1.73 \times 10^{3} g$ = 1500 g + mass of Styrofoam

(1730 - 1500) g = mass of Styrofoam

mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

Density = $\frac{mass}{volume}$

= $\frac{230}{4.32 \times 10^{3} cm^{3}}$

= $53.24 \times 10^{-3} g cm^{-3}$

Thus, we can conclude that the density of Styrofoam is $53.24 \times 10^{-3} g cm^{-3}$ .

4 0

3 years ago

Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f

Yanka [14]

The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

H2O + HCCH + 5/2 O=O 2CO2

Heat of reaction for acetylene combustion = 1259kj/mol

CO2 has a molecular mass of 44g/mol.

2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

Learn more about acetylene here :

brainly.com/question/15346128

#SPJ4

6 0

1 year ago

In this lab, you will use the flame test to identify the metal ion in two boxes of unidentified fireworks. Write an investigativ

Rasek [7]

In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis.

Following are the investigative question that you can answer by doing this experiment.
1) What information can be obtained from the colour of flame?
2) State the relationship between wavelength, frequency, and energy?
3) Can you identify the metal present in unknown sample provided?
4) How will you identify amount of metal present in sample solution?
5) Why do different chemicals emit light of different colour?

6 0

3 years ago

Read 2 more answers

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not

frutty [35]

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

A + B → C

the rate law can be expressed as:

Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

Rate₁ = k[A]₁ᵃ[B]ᵇ → eq. 1

Rate₂ = k[A]₂ᵃ[B]ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

(2) = (2)ᵃ
taking log of both sides
log (2) = a Log (2)
0.693 = a * 0.693
a =1

∴ order of reaction with respect to A is first (=1) → (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

Rate₁ = k[A]ᵃ[B]₁ᵇ → eq. 1

Rate₂ = k[A]ᵃ[B]₂ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

(1) = (2)ᵇ
taking log of both sides
log (1) = b Log (2)
0 = 0.693 * b
b = 0

∴ order of reaction with respect to B is zero → (2)

So, from 1 and 2 the right choice is rate = k[A]¹[B]⁰= k[A]

6 0

3 years ago