What effect does salt have on most metals?

Answers

Answer 1

Answer: the combination of moisture, oxygen and salt ,especially sodium chloride, damages metal worse than rust does. This combination corrodes, or eats away at, the metal, weakening it and causing it to fall apart.

Related Questions

A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached New England, the rain it produced had an average pH of 4.66. How much more acidic was the rain falling in New England?
The amount of heat required to melt 2 lbs of ice is twice the amount of heat required to melt 1 ib of ice. is this observation a macroscopic or microscopic description of chemical behavior? Explain your answer.
A compound has a molecular formular of C12H24O6.What is the compound's empirical formula
It is possible for gases in the atmosphere to change the behavior of energy on earth. Agree or disagree
What is the chemical formula for the ionic compound formed by Au3+ andHSO3-?

100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

Answers

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature changes from 5° C to 30° C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) The volume coefficient of expansion γγ for alcohol = 1.12 x 10-4 K-1

Answers

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

$v_0=500$
$\alpha =1.12* 10^(-4)$
$\Delta \epsilon = 25$

Now,

Increase in volume will be:

⇒ $\Delta V = \alpha* v_0* \Delta \epsilon$

By putting the given values, we get

$=1.12* 10^(-4)* 500* 25$

$=1.12* 10^(-4)* 12500$

$=1.4 \ mL$

Can aniline be nitrated directly?

Answers

Answer:

Yes.

Explanation:

Hello!

In this according to the attached file, we infer that the aniline can be nitrated by the addition of nitric acid and in presence of sulfuric acid that provides an acidic media. It leads to the formation of o-nitroaniline, m-nitroaniline and p-nitroaniline whereas the major products are the last two due to the steric hindrance.

Best regards!

What is the mass, in grams, of 1.50 mol of iron (III) sulfate?Express your answer using three significant figures.

Answers

Answer : The mass of 1.50 mole of iron(III) sulfate is, $5.99* 10^2g$

Explanation : Given,

Moles of iron(III) sulfate = 1.50 mole

Molar mass of iron(III) sulfate = 399.88 g/mole

The formula of iron(III) sulfate is, $Fe_2(SO_4)_3$

Formula used :

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3* \text{Molar mass of }Fe_2(SO_4)_3$

Now put all the given values in this formula, we get:

$\text{Mass of }Fe_2(SO_4)_3=1.50mole* 399.88g/mole$

$\text{Mass of }Fe_2(SO_4)_3=599.82g=5.99* 10^2g$

Therefore, the mass of 1.50 mole of iron(III) sulfate is, $5.99* 10^2g$

A 51.9g sample of iron, which has a specific heat capacity of 0.449·J·g?1°C?1, is put into a calorimeter (see sketch at right) that contains 300.0g of water. The temperature of the water starts off at 19.0°C. When the temperature of the water stops changing it's 20.3°C. The pressure remains constant at 1atm. Calculate the initial temperature of the iron sample. Be sure your answer is rounded to 2 significant digits.

Answers

Answer:

the initial temperature of the iron sample is Ti = 90,36 °C

Explanation:

Assuming the calorimeter has no heat loss to the surroundings:

Q w + Q iron = 0

Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )

Assuming Q= m*c*( T- Tir)

mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

Tir = 90.36 °C

Note :

- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C

- We assume no reaction between iron and water

Final answer:

To calculate the initial temperature of the iron sample, use the equation q = m * c * T, where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature which is 90.36 °C

Explanation:

To calculate the initial temperature of the iron sample, we can use the equation:

q = m * c * T

Where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, andT is the change in temperature. In this case, we know the mass of the iron sample, the specific heat capacity of iron, and the change in temperature of the water. By rearranging the equation, we can solve for the initial temperature of the iron sample.

Thus,

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

Tir = 90.36 °C

Learn more about Calculating initial temperature here:

brainly.com/question/30370638

#SPJ12

The value of the Solubility Product Constant for lead phosphate is ____________Write the reaction that corresponds to this Ksp value.

_______(Aq,S,L) +_______(Aq,S,L) <-------->_______(Aq,S,L) +_______(Aq,S,L)

Ksp values are found by clicking on the "Tables" link.

Use the pull-down menus to specify the state of each reactant or product.

If a box is not needed leave it blank.

Answers

Answer: The reaction for the $K_(sp)$ value of lead phosphate is given below and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is expressed as $K_(sp)$

The chemical formula of lead phosphate is $Pb_3(PO_4)_2$

The equation for the hydration of the lead phosphate is given as:

$Pb_3(PO_4)_2(s)+H_2O(l)\rightarrow 3Pb^(2+)(aq.)+2PO_4^(3-)(aq.)$

The solubility product of lead phosphate is $3.0\rightarrow 10^(-44)$ . This means that it is highly insoluble in water as the solubility product is very very low.

Hence, the reaction for the $K_(sp)$ value of lead phosphate is given above and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Other Questions

Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature

What is the number of neutrons in Hydrogen, O (A= 1, Z= 1)?

Show the conversion factor from Patolbf/ft2is 0.02089.

Nickel (II) ions form a complex ion in the presence of ammonia with a formation constant (Kf) of 2.0×10^8:Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ Calculate the molar solubility of NiS in 3.1 M NH3. g