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Orlov [11]
3 years ago
11

It is possiblefor an object's weight to change while its mass remains constan?Explain.

Chemistry
1 answer:
Georgia [21]3 years ago
7 0

Answer:

Yes. Weight is the product of mass times gravitational acceleration. So all you have to do is vary the gravitational field and you vary weight.

Explanation:

You might be interested in
What is the mass of 5 moles of Fe2(CO3)3 ?
sineoko [7]

Answer:

1218.585

Explanation:

Looking at the subscripts we know there are 2 atoms of Fe, 3 atoms of C, and 6 of O.

Take the molar mass of each atom (from the periodic table) and multiply by the # of atoms

Fe: 55.845×2= 111.69

C: 12.011×3= 36.033

O:15.999×6=95.994

Add the values together: 243.717 g/mol

That is 1 mole of the molecule. Multiply by 5 for the final answer.

243.717×5=1218.585

7 0
2 years ago
What is the energy change if 84.0 g of calcium oxide (CaO) reacts with excess water in the following reaction?
pochemuha
   The energy change  if 84.0 g   of CaO  react  with  excess  water is  98KJ of heat is released.

calculation
heat  =  number of moles  x  delta H

delta H = - 65.2  Kj/mol

first find the number of  moles  of  CaO reacted

moles = mass/molar mass
the molar mass  of CaO =  40 +  16=  56 g/mol
mass = 84 g
moles therefore = 84 g/56 g/mol =1.5 moles

Heat is therefore =  1.5 moles  x -65.2 = - 97.8 Kj = -98 Kj

  since  sign is  negative  the   energy  is released 

6 0
3 years ago
Determine whether you can swim in 1.00 x 10^27 molecules of water.​
zloy xaker [14]

Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

7 0
2 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
2. When a teaspoonful of sugar is added to water in a beaker,
Ket [755]
Sugar + water = 2) a solution
3 0
3 years ago
Read 2 more answers
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