List the properties of copper, sulfur and copper sulfide?

Most elements will have the same number as what?

MrRissso [65]

Answer:

An element's atomic number is equal to the number of protons in the nuclei of any of its atoms. The mass number of an atom is the sum of the protons and neutrons in the atom. Isotopes are atoms of the same element (same number of protons) that have different numbers of neutrons in their atomic nuclei

3 0

2 years ago

2. Energy cost of plasma generation of NO, NO2, and NO3 in air (or air bubble) is about 20 eV per N-atom. 100 W plasma system ha

svet-max [94.6K]

Answer:

0.0933 moles/Litre

Explanation:

We assume that the number of moles of N- used is equal to the number of moles of Nitrogen containing compounds that are generated due to the fact that the nitrogen containing compound that are produced contain only one nitrogen in each atom. As such, finding the amount of nitrogen used up explains the amount of compound formed. This can be expressed as follows:

Energy cost = $\frac{20ev}{N- atom}$

Given that:

Energy = 100 W for 60 minutes

100 W = 100 J/s

= 100 J/s × (60 × 60) seconds

= 3.6 × 10⁵ J

Let now convert 3.6 × 10⁵ J to eV; we have:

= ( 3.6 × 10⁵ × 6.242 × 10¹⁸ )eV

= 2.247 × 10²⁴ eV

So, number of N-atom used up to form compounds will now be:

= 2.247 × 10²⁴ eV × $\frac{1}{20eV} N-atom$

= 1.123 × 10²³ N-atom

To moles; we have:

= $\frac {1.123*10^{23}} {6.023*10^{23}}$

= 0.186 moles

However, we are expected to leave our answer in concentration (i.e in moles/L)

since we are given 2L

So; 0.186 moles ⇒ $\frac{0.186 moles}{2 Litre}$

= 0.0933 moles/Litre

3 0

2 years ago

An airplane has a mass of 12,000kg. If it is accelerating at 16m/sec2, what is its force?(round to the nearest whole number) NEE

icang [17]

Answer:

$F=1.92x10^5N$

Explanation:

Hello!

In this case, since the force exerted by an object with a certain mass an acceleration is:

$F=m*a$

We can plug in the given mass of 12,000 kg and acceleration of 16 m/s^2 to obtain:

$F=12,000kg*16m/s^2\\\\F=192,000kg*m/s^2\\\\F=1.92x10^5N$

Best regards!

8 0

2 years ago

Question 044 Which of the following sequences converts 2-methylpropene and sodium acetylide into 3-methylbutanal? 1) HBr; 2) NaC

Minchanka [31]

Answer:

1) HBr; 2) NaCCH; 3) O3; 4) H2O

Explanation:

The first step is formation of alkyl halide followed by reaction with sodium acetylide, to form 3-methylbutene, which is then followed by oxidation reaction with O3& H2O to 3-methylbutanal

8 0

3 years ago

Read 2 more answers

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

2 years ago