25 POINTS:)

Consider the reaction, 2 d(g) + 3 e(g) + f(g) => 2 g(g) + h(g) when h is increasing at 0.64 mol/ls, how quickly is e decreasi

bekas [8.4K]

Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → 2G(g) + H(g).
H is increasing at 0,64 mol/L·s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
E is decreasing at 1,92 mol/L·s.

8 0

3 years ago

What are the molarity and the normality of a solution made

Nikitich [7]

Explanation:

It is known that molarity is the number of moles present in a liter of solution.

Mathematically, Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

Hence, calculate the molarity of given solution as follows.

Molarity of citric acid = $\frac{\text{mass of citric acid}}{\text{molar mass of citric acid}} \times \frac{1}{\text{volume of solution(L)}}$

= $\frac{25 g}{192.13 g/mol} \times \frac{1}{0.75 L}$

= 0.173 M

As citric acid is a triprotic acid so, upon dissociation it gives three hydrogen ions.

Normality = Molarity × no. of hydrogen or hydroxide ions

= 0.173 × 3

= 0.519 N

Thus, we can conclude that molarity of given solution is 0.173 and its normality is 0.519 N.

3 0

3 years ago

Write a question here it’s simple.

dusya [7]

Answer:

what type of question?

Explanation:

whats your favorite color???

3 0

3 years ago

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

Answer: The specific heat of metal is 0.821 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

8 0

3 years ago

19. I accidentally mixed bleach and ammonia when cleaning one day. It released 55 liters of deadly chlorine (CI) gas. How many g

user100 [1]

55,000 grams
Hope this helps you out :)

3 0

3 years ago