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3 years ago

What is the symbol for the most common isotope of tin

Chemistry

2 answers:

ANTONII [103]3 years ago

8 0

Answer:

The symbol "Sn" comes from the Latin word for tin, "stannum." Tin has ten stable isotopes. This is the most stable isotopes of all the elements. The most abundant isotope is tin-120.Tin is too expensive. Also, tin does not 'rust', although it oxidizes. Your rust is iron oxide. Galvanized steel is steel with a thin zinc coating, likely hot-dip galvanization.

Explanation:

The symbol Sn for tin is an abbreviation of the Latin word for tin, stannumIsotopes of tin. Tin (50Sn) is the element with the greatest number of stable isotopes (ten; three of them are potentially radioactive but have not been observed to decay), which is probably related to the fact that 50 is a "magic number" of protons.Uses and properties

Below 13°C it slowly changes to a powder form. Tin has many uses. It takes a high polish and is used to coat other metals to prevent corrosion, such as in tin cans, which are made of tin-coated steel. Alloys of tin are important, such as soft solder, pewter, bronze and phosphor bronze.

Hope This Helps Have A Nice Day :) !!

erastovalidia [21]3 years ago

3 0

Answer Sn

Explanation:

The symbol "Sn" comes from the Latin word for tin, "stannum." Tin has ten stable isotopes. This is the most stable isotopes of all the elements. The most abundant isotope is tin-120.

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How many valence electrons are in potassium

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3 years ago

1- A sample of gas is in a 3.00 L contain at a pressure of 740.0 mmHg. What is the new pressure of the sample if the container's

inna [77]

Considering the Boyle's law, the new pressure of the sample is 1,776 mmHg.

<h3>What is Boyle's law</h3>

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

Boyle's law states that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

Now it is possible to assume that you have a certain volume of gas V1 which is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and the following will be true:

P1×V1=P2×V2

<h3>New pressure</h3>

In this case, you know:

P1= 740 mmHg
V1= 3 L
P2= ?
V2= 1.25 L

Replacing in Boyle's law:

740 mmHg× 3 L=P2× 1.25 L

Solving:

P2= (740 mmHg× 3 L) ÷ 1.25 L

P2= 1,776 mmHg

Finally, the new pressure of the sample is 1,776 mmHg.

Learn more about Boyle's law:

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7 0

2 years ago

Read 2 more answers

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

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