C6h12(l)+9O2(g)->6CO2(g)+6h2O(g)

A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and

kvasek [131]

Given:

Moles of He = 15

Moles of N2 = 5

Pressure (P) = 1.01 atm

Temperature (T) = 300 K

To determine:

The volume (V) of the balloon

Explanation:

From the ideal gas law:

PV = nRT

where P = pressure of the gas

V = volume

n = number of moles of the gas

T = temperature

R = gas constant = 0.0821 L-atm/mol-K

In this case we have:-

n(total) = 15 + 5 = 20 moles

P = 1.01 atm and T = 300K

V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L

Ans: Volume of the balloon is around 488 L

3 0

3 years ago

Hypothesis and Data Collection Part A The image shows three different liquids: water, water with salt, and vinegar. The pH of ea

Norma-Jean [14]

Answer:

The water would be neutral, (usually 7). The salt water would be the same (7) and the vinegar would be very acidic. (probably 2).

Explanation:

7 0

1 year ago

What is the % composition of Carbon in Chromium (iii) Carbonate

photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

$Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3$

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

$\begin{gathered} C\rightarrow3\times12=36 \\ \\ O\rightarrow9\times16=144 \\ \\ Cr\rightarrow2\times52=104 \end{gathered}$

The molar mass will be thus:

$M=36+104+144=284\text{ g/mol}$

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

$\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\ \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}$

The percent composition of Carbon is thus 12.7 %.

8 0

1 year ago

Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere

Colt1911 [192]

Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
    4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
  = 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
  0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M

8 0

3 years ago

.

Debora [2.8K]

Explanation:

Their force of attraction is strongest in the middle of the magnet.

4 0

2 years ago

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