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3 years ago

20 points and brainliest for best answer

2 answers:

8 0

You can find the atomic mass of each on any standard periodic table.
Gold (Au): 196.97
Potassium (K): 39.098
Francium (Fr): 223
Copper (Cu): 63.546
Bromine (Br): 79.904
Arsenic (As): 74.992
Uranium (U): 238.03
Chromium (Cr): 51.996
Mercury(Hg): 200.59
Hydrogen (H): 1.008

~I hope this helps!~

RSB [31]3 years ago

7 0

Gold/Atomic mass

196.96657 u ± 0.000004 u

Potassium/Atomic mass

39.0983 u ± 0.0001 u

Francium/Atomic number

87

Copper/Atomic mass

63.546 u ± 0.003 u

Bromine/Atomic mass

79.904 u ± 0.001 u

Arsenic/Atomic number

33

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A 7.83 g sample of hcn contains 0.290 g of h and 4.06 g of n. find the mass of carbon in a sample of hcn with a mass of 3.37 g.

xeze [42]

To obey the Law of Conservation of Mass, the sum of all individual elements of a compound is equal to the mass of the compound. So, if HCN has a mass of 7.83 grams, then

7.83 g = mass of H + mass of C + mass of N

We know the masses of H and N to be 0.290 g and 4.06 g, respectively. Hence, we can find for the mass of C:

7.83 = 0.29 + mass of C + 4.06
mass of C = 3.48 g

As an extension to the Law of Conservation of Mass, there is also a Law of Definite Proportions. According to Dalton's atomic theory, a compound is formed from a fixed ratio of its individual elements. From our previous calculations, we know that the mass ratio of H to C to N is 0.29 g: 3.48 g:4.06 grams. The ratio could also be expressed in percentages. Let's find the mass percentage of Carbon in HCN to be used later:

mass % of Carbon = (3.48 g/7.83 g)*100
mass % of Carbon = 44.44%

So, if you collect a different mass of HCN, say 3.37 g, the corresponding mass of Carbon is equal to:

Mass of Carbon = (3.37)(44.44%)
Mass of Carbon = 1.498 g

3 0

3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

3 years ago

Student have a receipt which calls for 1.15 lb of flour, but, of course, in France you will need to purchase flour in kilograms.

Verizon [17]

0.52164 kg of Flour

<u>Explanation:</u>

We have to convert lb (pounds) kg (kilograms), by using the conversion unit,

1 lb = 0.4536 kg

So to convert 1.15 lb into kg, we have to multiply 1.15 by 0.4536.

$1.15 \times 0.4536$ = 0.52164 kg of Flour

So the receipt calls for 0.52164 kg of Flour.

5 0

3 years ago

Explain viscosity <br> Pls help ty

Fantom [35]

The viscosity of a fluid is a measure of its resistance to deformation at a given rate. For liquids, it corresponds to the informal concept of "thickness": for example, syrup has a higher viscosity than water.

3 0

2 years ago

Read 2 more answers

What mass of carbon monooxide must be burned to produce 175 kJ of heat under standard state condaitions?

Sphinxa [80]

Answer:

17.3124 grams

Explanation:

Given;

Amount of heat to be produced = 175 kJ

Molar mass of the carbon monoxide = 12 + 16 = 28 grams

Now,

The standard molar enthalpy of carbon monoxide = 283 kJ/mol

Thus,

To produce 175 kJ heat, number of moles of CO required will be

= Amount heat to be produced / standard molar enthalpy of CO

= 175 / 283

= 0.6183

Also,

number of moles = Mass / Molar mass

therefore,

0.6183 = Mass / 28

Mass of the CO required = 0.6183 × 28 = 17.3124 grams

7 0

3 years ago