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jekas [21]
3 years ago
5

What is the strength of alloys and what is the solubility of alloys?

Chemistry
1 answer:
il63 [147K]3 years ago
3 0

Answer:

The most common reason for alloying is to increase the strength of a metal. This requires that barriers to slip be distributed uniformly throughout the crystalline grains. On the finest scale, this is done by dissolving alloying agents in the metal matrix (a procedure known as solid solution hardening

Explanation:

hey why u search i have book that answer i got mark as brainlist please okkkkkk

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Characteristics of a medium wave
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4 0
3 years ago
Read 2 more answers
ASAP!
Andrei [34K]

Answer:

8.3334%

Explanation:

You have two masses. To find the percent of sodium chloride in water by mass, you divide the mass of NaCl by water. First, make both units the same. Easiest is to convert kg into g. 1.5kg = 1500g

125g NaCl/1500g H2O = 0.0833333333 ==> 8.3334%

5 0
3 years ago
How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be usefu
In-s [12.5K]

Answer:

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.

q 1= m c (T2-T1)

q1 = 36.0 g (4.18 J/gC) (100 - 65 C)

q1 = 5267 J =5.267kJ

next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2= 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3= 36.0 g (2.01 J/gC)(115-100C)

q3 = 1085 J =1.085kJ

qnet=q1 +q2 +q3

energy required=qnet=87.75kJ

8 0
3 years ago
93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP
Dahasolnce [82]
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
V= 93ml  to  liters =  93/1000= 0.093L
T=  10  +  273.15 = 283.15k

n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

At  STp    1  mole  =  22.4L
what about  3.72  x10^-3 moles

by  cross  multiplication

volume =  (3.72  x10^-3)mole  x  22.4L/  1  moles  = 0.083 L   or  83.3 Ml
3 0
3 years ago
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