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Korvikt [17]
2 years ago
13

A sample of bismuth has a mass of 343g and volume of 35.0 cm3. what is the density of bismuth

Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
4 0
I don’t think I could answer this sorry.........
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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.020 h, and the assem
nasty-shy [4]

Answer: The workdone W = 505J

Explanation:

Applying the pressure-volume relationship

W= - PΔV

Where negative sign indicates the power is being delivered to the surrounding

W = - 1.0atm * ( 5.88 - 0.9)L

= - 1.0atm * (4.98)

W = -4.98 atmL

Converting to Joules

1atmL = 101.325J

-4.98atmL = x joules.

Work done in J = -4.98 * 101.325

W= -505J

Therefore the workdone is -505J

5 0
2 years ago
a gas sample is heated from -20.0 C to 57.0 C and the volume is increased from 2.00L to 4.50L. If the initial pressure is 0.109a
Andreas93 [3]
Idk what this is im not good at this 
7 0
2 years ago
22. How many atoms are there in 344.75 g of gold nugget? a. 1.05 x 10 to the power of 24 atoms b. 1.05 x 10 to the power of 23 a
mixas84 [53]

Answer:

1.053×10²⁴ atoms of gold

Explanation:

Hello,

Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.

In this question, we're required to find the number of atoms in 344.75g of a gold nugget.

We can use mole concept relationship between Avogadro's number and molar mass.

1 mole = molar mass

Molar mass of gold = 197 g/mol

1 mole = Avogadro's number = 6.022 × 10²³ atoms

Number of mole = mass / molar mass

Mass = number of mole × molar mass

Mass = 1 × 197

Mass = 197g

197g is present in 6.022×10²³ atoms

344.75g will contain x atoms

x = (344.75 × 6.022×10²³) / 197

X = 1.053×10²⁴ atoms

Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold

5 0
2 years ago
How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?
mash [69]

Answer : Both solutions contain 3.011 X 10^{23} molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain 3.011 x 10^{23} molecules.

Avogadro's Number is  N_{A} =  6.022 X 10^{23} which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules  = N_{A} X M;

∴  Number of molecules =  6.022 X 10^{23} X 0.5 mol/L X 1 L which will be  = 3.011 X 10^{23}

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

7 0
2 years ago
Read 2 more answers
5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is
GenaCL600 [577]
1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V  = 7.9 atm
3 0
2 years ago
Read 2 more answers
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