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statuscvo [17]
3 years ago
10

Look at the image below. What is the ratio of the formula shown? *

Chemistry
2 answers:
tamaranim1 [39]3 years ago
8 0
It’s the 2 one lmk if it’s correct
aleksklad [387]3 years ago
6 0
It’s 1:1 b3cuae that’s the one u can
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A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m
Lana71 [14]

Answer:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol

Moles of hydrogen gas produced = 0.01225 mol

2M+2xHCl\rightarrow 2MCl_x+xH_2

Moles of metal =\frac{0.225 g}{27.0 g/mol}=8.3333 mol

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

\frac{8.3333}{0.01225 mol}=\frac{2}{x}

x = 2.9 ≈ 3

2M+6HCl\rightarrow 2MCl_3+3H_2

MCl_3\rightarrow M^{3+}+Cl^-

Formulas for the oxide and sulfate of M will be:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3.

3 0
4 years ago
8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is
Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

T^2\propto R^3

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3

(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3  

\dfrac{1}{64}=(\dfrac{R}{R'})^3

R'=4\times R

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.  

4 0
3 years ago
What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?
Elden [556K]

The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.  

<h3>What is pH ?</h3>

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻   (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

<em>pH = 14 - pOH</em>

     =14 - [-log[OH⁻]]

    = 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

0.64 - x                          x              x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹  =[C₅H₅NH⁺]  [OH⁻]  /  [C₅H₅N]

                = x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

  • x₁ = -3.32 x 10⁻⁵
  • x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

     = 14 + log (3.32 x 10⁻⁵)

 

     = 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

brainly.com/question/8834103?referrer=searchResults

#SPJ1

3 0
2 years ago
what is the percent by mass of NaHCO3 in a solution containing 20.0 g of NaHCO3 dissolved in 600.0 ml of water
Ne4ueva [31]
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%

m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water

%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
7 0
3 years ago
I need help please, will give 5 stars
DedPeter [7]
2nd one bhutdsaadxjytwwdghurfc
5 0
3 years ago
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