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Alex
3 years ago
6

When a 10 ml graduated cylinder is filled to the 10 ml mark, the mass of the water was measured to be 9.955 g. if the density of

water is taken to be 0.9975 g/ml, what is the percent error for the 10 ml of water?
Chemistry
1 answer:
Lady_Fox [76]3 years ago
5 0

Given, the density of water is  0.9975 g/ml. Density of water is mass of water per unit volume. Mass of 1 ml of water supposed to be  0.9975 g from density of water. So, mass of 10 ml of water is (0.9975 X 10) g= 9.975 g. From graduated cylinder, mass of 10 ml water is measured to be 9.955 g. So, error for mass of 10 ml water= (9.975-9.955)=0.02 g. Percentage of error for 10 ml water is \frac{0.02}{10} X 100 = 0.2. Error in the mass  for the 10 ml of water is 0.2 %.

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Explain what a mole of copper (Cu) atoms and a mole of Sulfur Dioxide (SO2) molecules have in common.
evablogger [386]

Answer:

The amount of molecules → 6.02×10²³

Explanation:

Avogadro Number (NA) is the amount of particles that are contained in 1 mol of anything.

It does not matter the mass of compounds, or molecules, 1 mol has 6.02×10²³ particles, the same number for every compound.

1 mol of H₂O that weighs 18 grams has 6.02×10²³ molecules

1 mol of sucrose, that weighs 342 g has 6.02×10²³ molecules

Generally, the NA  is the number of atoms in 12 grams of the carbon-12 isotope.

6 0
3 years ago
Complete the sentences to identify the trend in first ionization energies as one proceeds down the group 7A elements and explain
Harrizon [31]

Answer:

a)decrease

b)increase

c)smaller

d)smaller

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3 0
3 years ago
If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r
kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)

Energy absorbed by pork,E = 1.6\times 10^3 kJ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

10\%=\frac{E}{Q}\times 100

Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = -1.6\times 10^4 kJ

Moles of propane burnt to produce Q energy =n

n\times \Delta H_{rxn}=Q

n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

\frac{1}{3}\times 7.83 mol=23.49 mol carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0
3 years ago
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