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3 years ago

Help if possible! Picture attached with the problems.

Chemistry

1 answer:

MAVERICK [17]3 years ago

8 0

1. 1 M , 2 M , 1 M
2. 10 mol , 0.1 mol , 0.5 mol
3. 0.5 L , 6.6 L , 5/21 L

M = mol/L

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What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques?

amid [387]

Answer:

The principal elements are Iron, Nickel, Chromium

It has 0.37% to 0.43% Carbon

It is hardenable.

It can be hardened by cold working, annealing, quenching.

Explanation:

The chemical composition of AISI 4340 Steel are as follows:

Iron --- 95% to 96%

Nickel --- 1.6% to 2.0%

Chromium --- 0.7% to 0.9%

Manganese --- 0.6% to 0.8%

Carbon --- 0.37% to 0.43%

Molybdenum --- 0.2% to 0.3%

Silicon --- 0.15% to 0.30%

Sulfur --- 0.040%

Phosphorous --- 0.0350%

So, the principal alloy elements from this composition are Iron, Nickel and Chromium

The carbon content in this alloy is 0.37% to 0.43%

Yes, it can be hardened.

It can be Hardened by Cold Working, Annealing or Quenching.

3 0

3 years ago

What is the mass number of an iron atom that has 29 neutrons?

hjlf

Answer:

The answer is 54

Explanation:

8 0

3 years ago

The bond between sodium and chlorine in the compound sodium chloride (NaCl) is a/an:

Deffense [45]

The classic case of ionic bonding, the sodium chloride molecule forms by the ionization of sodium and chlorine atoms and the attraction of the resulting ions. An atom of sodium has one 3s electron outside a closed shell, and it takes only 5.14 electron volts of energy to remove that electron.

8 0

3 years ago

Please help me.. what is carbon 1/12

tatuchka [14]

Answer:

It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units. ... In this scale 1 atomic mass unit (amu) corresponds to 1.660539040 × 10−24 gram.

6 0

3 years ago

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

4 years ago