Name the following straight-chain

copper forms two oxides. On heating 1 g of each in hydrogen 0.888 g and 0.798 g of the metal was obtained. Show that the results

Mashutka [201]

Answer:

For the first oxide, 1 g gives 0.888 g of copper.

Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.

For the second oxide, 1 g gives 0.798 g of copper.

Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.

So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.

Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.

6 0

2 years ago

Read 2 more answers

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago

Diana raises a 1000. N piano a distance of 5.00 m using a set of pulleys. She pulls

Alina [70]

Answer:

a) 250 N

b) 50 N

c) 5000 J

d) 4

e) 3.33

Explanation:

Given that weight of piano ( $F_r$ ) = 1000 N, distance moved by pulley ( $d_r$ ) = 5 m and the rope used ( $d_e$ ) =20 m

a) The effort applied ( $F_e$ ) if it was an ideal machine is:

$F_e=\frac{F_rd_r}{d_e} = \frac{1000*5}{20}=250N$

b) If the actual effort = 300 N

The force to overcome friction = actual effort - $F_e$ = 300 - 250 = 50 N

c) Work output = $F_rd_r=1000*5=5000J$

d) ideal mechanical advantage (IMA) = $\frac{d_e}{d_r} =\frac{20}{5} = 4$

e) Input force = 300 N, Therefore:

actual mechanical advantage = $F_r$ / input force = 1000 / 300 = 3.33

5 0

3 years ago

Gases do not settle at the bottom of the vessel why

bulgar [2K]

Explanation:

Gases are very less denser also, they've negligible intermolecular force of attraction between the particles of the gas. So, they all are free to roam seperately and hence making a negligible volume for which they become heavy settle down.

7 0

2 years ago

In a beaker, 300 grams of water at 80°C cools down to 20°C. Assume all the heat from the water is absorbed by 100 cubic meters o

scZoUnD [109]

Answer:

mass of water, mw = 300g = 0.3kg

∆Tw = (80 - 20) °C

volume of air, va = 100m³

mass of air, ma = 100g = 0.1kg

∆Ta = ?

H = mc∆T

Hw = mwcw∆Tw

Hw = 0.3*4200*60

Hw = 75600J

Hw = 75.6 kJ

All the above heat energy got absorbed by air,

that is; Ha = 75600J

since it's given that the heat was absorbed by a specific amount of volume of air

then specific capacity of volume of air is

then,

ca = Ha × density

ma temp

then,

Ha = vaca∆Ta

where, ca = volumetric heat capacity of air = 0.012kJ/m³°C

75.6kJ = 100m³ × 0.012kJ/m³°C × ∆Ta

75.6 = 1.2/°C × ∆Ta

∆Ta = 63°C

63°C is the temperature change in air.

4 0

2 years ago