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3 years ago

Formula for tin (II) sulfide

Chemistry

2 answers:

irakobra [83]3 years ago

5 0

Answer:

SnS

Explanation:

Tin, as it shows on the periodic table can have a +2 or +4 charge.

Sulfide is just Sulfur which has a -2 charge.

It tells us the charge of the tin in roman numerals. In this case (II) means two, so it is tin with a charge of two.

In order for these to balance each other, we are going to take one of the tin and one of the sulfur because it the +2 and -2 add to zero.

igor_vitrenko [27]3 years ago

4 0

Answer:

SnS

Explanation:

Tin(II) sulfide is a chemical compound of tin and sulfur. The chemical formula is SnS. Its natural occurrence concerns herzenbergite, a rare mineral.

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The effective nuclear charge for an atom is less than the actual nuclear charge due to

Irina18 [472]

Answer:

A = shielding.

Explanation:

The addition of electron causes the atomic size increase from top to bottom due to increase in atomic number.

As the atomic number increased one more electron is added and because of this electron on more electronic shell is added. Thus the electron become more away from the nucleus as many of other electrons are present in the way from nucleus to the outer electrons.

The hold of nucleus becomes weaker. Although nuclear charge is also increased but at the same time other electrons shield the respective electrons. So effective nuclear charge is weaker than the actual nuclear charge.

Because of this shielding it is easy to remove the electrons or we can say ionization energy decreases.

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4 years ago

Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m

Nataly_w [17]

The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

Trial [A] [B] Rate 
 (M) (M) (M/s) 
1 0.50 0.010 3.0×10−3 
2 0.50 0.020 6.0×10−3 
3 1.00 0 .010 1.2×10−2

Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0 =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s

8 0

3 years ago

Read 2 more answers

The green light emitted by a stoplight has a wavelength of 525 nm. What is the frequency of this photon? (c = 3.00 × 10⁸ m/s).

Reil [10]

The frequency of this photon is 5.94 x $10^{14} H_{Z}$ .

Solution:

The speed of light has a value of approximately 3.00×108 m/s

The wavelength should have units of meters.

The frequency should have units of Hz or s−1. Hz is hertz or reciprocal seconds.

We know the value of the speed of light and we are given the wavelength. All we have to do is rearrange the equation to solve for the frequency:

v = c/λ

= 3.00× $10^{8}$ m/s/505× $10^{-9}$ m

= 5.94X $10^{14}$ s-1

= 5.94X $10^{14}$ Hz

Frequency describes the number of waves passing through a particular location at a particular time. A class interval frequency is the number of observations that occur in a particular predefined interval. So, for example, if 20 of her ages 5-9 appear in the survey's data, the interval 5-9 has a frequency of 20. The class interval endpoints are the lowest and highest possible values of the variable.

Learn more about The frequency here:- brainly.com/question/254161

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1 year ago

If a liter of water is heated from 20c to 50c what happens to its volume

aleksklad [387]

The volume increases to 1.009 L.

V= V_0 +βΔT

The thermal expansion coefficient (β) of water changes with temperature, so we must calculate the volume change over small (10 °C) intervals.

20 °C to 30 °C: V = 1 L + 0.000 207 L·°C^(-1) × 10 °C = 1.002 07 L

30 °C to 40 °C: V = 1.002 07 L + 0.000 303 L·°C^(-1)] × 10 °C = 1.005 10 L

40 °C to 50 °C: V = 1.005 10 L + 0.000 385 L·°C^(-1)] × 10 °C = 1.008 95 L

The volume increases by about 9 mL when the temperature increases from 20 °C to 50 °C.

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4 years ago

Atoms of oxygen have a total of 8 electrons. Are these atoms stable, and why or why not? a.) Yes, because the Octet Rule says st

garik1379 [7]

D) otherwise we wouldn't be able to have water, oxygen gas, and all other compounds that have oxygen. We need at least 1 free slot to make a molecule

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4 years ago