A metal can form

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

3 years ago

What is the name of Cr3P

almond37 [142]

Answer:

Chromium (I) phosphide.

Explanation:

Hope this help:D

8 0

3 years ago

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

blondinia [14]

Specific heat is the amount of heat absorb or released by a substance to change the temperature to one degree Celsius. To determine the specific heat, we use the expression for the heat absorbed by the system. Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)

By substituting the given values, we can calculate for C which is the specific heat of the material.

2510 J = .158 kg ( 1000 g / 1 kg) (C) ( 61.0 - 32.0 °C)
C = 0.5478 J / g °C

6 0

3 years ago

The diagram shows monochromatic light passing through

Maru [420]

Destructive interference in which waves cancel each other out is depicted in region X,Y and Z.

Explanation:

Interface is the particle property of light waves. When incident light beam is made to pass through holes, the waves will combine either constructively or destructively. Constructive interference means the waves having same phase will get added so they will increase in amplitude. While destructive interference means the waves combining have different phases like crests and troughs. So they undergo decrease or complete vanishing of amplitude.

When waves combine in constructive interference, they form bright white light and when they combine in destructive interference, they form dark black light. So the regions X, Y and Z are shown as dark black colors in the diagram, so these regions represent destructive interference in which waves cancel each other out.

8 0

3 years ago

You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00

Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ $\frac{[Ac^-]}{[Acac]}$

Where Ac⁻ is the salt of acetic acid (Acac).

Solving:

1,738 = $\frac{[Ac^-]}{[Acac]}$ (1)

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] (2)

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = 0,0365 mol

Ac⁻ = 0,127 M × 0,500 L = 0,0635 mol

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × $\frac{60,05 g}{1 mol}$ = 6,005 g

You need to weight 6,005 g of acetic acid

I hope it helps!

7 0

4 years ago