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Vladimir79 [104]

2 years ago

9

Approximately how much greater in mass is a d2o molecule than an h2o molecule? (assume the oxygen atoms have a mass number of 16

. deuterium, d, is an isotope of hydrogen with a mass number of 2, 2h.)

2 answers:

faust18 [17]2 years ago

6 0

The deuteriuoted water that is $D_{2}O$ consists of an isotope of hydrogen in place of hydrogen from the normal water. The approximate mass of deuteriuoted water and normal water can be calculated as {(2×2)+16} = 20. However the exact mass of $D_{2}O$ is 20.027 g/mole, whereas the mass of water can also be calculated as- {(1×2)+16}=18, however the exact mass of ( $H_{2}O$ ) molecule is 18.015 g/mole. Thus the difference between the mass of the molecules are (20.027 - 18.015) = 2.01 2g/mole.

7nadin3 [17]2 years ago

3 0

Answer:

$D_2O$ molecule is greater than $H_2O$ molecule by 2 amu.

Explanation:

Mass of $H_2O$ molecule:

$M_{H_2O}$ = 2 × 1 amu + 1 × 16 amu = 18 amu

Mass of $D_2O$ molecule:

$M_{D_2O}$ = 2 × 2 amu + 1 × 16 amu = 20 amu

Difference between mass of $D_2O \& H_2O$

$M_{D_2O}-M_{H_2O} =20 amu - 18 amu = 2 amu$

$M_{D_2O} =2+M_{H_2O}$

$D_2O$ molecule is greater than $H_2O$ molecule by 2 amu.

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A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

2 years ago

Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1

pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0

2 years ago

You wish to make a buffer with pH 7.0. You combine 0.060 grams of acetic acid and 14.59 grams of sodium acetate and add water to

aleksandr82 [10.1K]

Answer:

The pH of the buffer is 7.0 and this pH is not useful to pH 7.0

Explanation:

The pH of a buffer is obtained by using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pH is the pH of the buffer</em>

<em>The pKa of acetic acid is 4.74.</em>

<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>

<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>

<em />

Replacing:

pH = 4.74 + log [0.1779mol] / [0.001mol]

<em>pH = 6.99 ≈ 7.0</em>

<em />

The pH of the buffer is 7.0

But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,

this pH is not useful to pH 7.0

<em />

7 0

2 years ago

Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod

son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0

2 years ago

One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng

Sav [38]

Answer:

8 Silicon atom are present in unit cell.

16 oxygen atoms are present unit cell.

Explanation:

Number of atoms in unit cell = Z =?

Density of silica = tex]2.32 g/cm^3[/tex]

Edge length of cubic unit cell = a = 0.700 nm = $0.700\times 10^{-7} cm$

$1 nm=10^{-7} cm$

Molar mass of Silica = $28.09 g/mol+16.00\times 2=60.09 g/mol$

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$2.32 g/cm3=\frac{Z\times 60.09 g/mol}{6.022\times 10^{23} mol^{-1}\times (0.700\times 10^{-7}cm)^{3}}$

$Z = 8$

1 silicon is 2 oxygen atoms. then 8 silicon atoms will be 16 oxygen atoms.

5 0

2 years ago