Hi can someone help me with these 2 :)

A gas is produced when 25 g of substance X is mixed into 60 g of solution Z in a beaker. The mass of the solution in the beaker

Zepler [3.9K]

mass of the gas = 10 g

<h3>Further explanation</h3>

<em>The law of conservation of mass :</em>

<em>In a closed / isolated system, the total mass of the substance before the reaction will be equal to the total mass of the reaction product.</em>

Reaction

$\tt X+Z\rightarrow gas+solution$

mass before reaction=mass after reaction

$\tt 25+60=gas+75\\\\gas=25+60-75=10~g$

5 0

3 years ago

Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces

inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>

aX → bY (1)

$rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}$

<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>

(a) Based on the definition above, we can express the rate of reaction (2) as follows:

3O₂(g) → 2O₃(g) (2)

$rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}$ (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:

$rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}$

$+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}$

$\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}$

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

Have a nice day!

6 0

4 years ago

It takes 42.25 ml of 0.0500 M Na2S2O3 solution to completely react with the iodine present in a 150.0 ml iodine solution. How ma

svlad2 [7]

Answer:

There were 0.268 grams I2 present

Explanation:

Step 1: Data given

Volume of Na2SO3 = 42.25 mL = 0.04225 L

Molarity of Na2SO3 = 0.0500 M

Volume of iodine = 150.0 mL

Step 2: The balanced equation

I2 + 2 Na2SO3 → Na2S2O6 + 2 NaI

Step 3: Calculate moles of Na2SO3

Moles Na2SO3 = molarity * volume

Moles Na2SO3 = 0.0500 M * 0.04225 L

Moles Na2SO3 = 0.0021125 moles

Step 4: Calculate moles I2

For 1 mol I2 we need 2 moles Na2SO3

For 0.0021125 moles Na2SO3 we have 0.00105625

Step 5: Calculate mass of I2

Mass I2 = moles * molar mass

Mass I2 = 0.00105625 * 253.8 g/mol

Mass I2 = 0.268 grams

There were 0.268 grams I2 present

3 0

4 years ago

Read 2 more answers

Which change in the state of matter is thermal energy released

ycow [4]

Answer:

Condensation

Explanation:

Thermal energy is released in this process

3 0

3 years ago

Read 2 more answers

Convert 100 degrees Celsius to Fahrenheit

SIZIF [17.4K]

Answer:

212 degrees Fahrenheit

5 0

4 years ago