How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl?

Hello can you please help me to solve above fill in the blank questions

LiRa [457]

Answer:

C2H5OH is a formula of ethyl alchol or ethanol.

alkenes are characterisized by carbon-carbon double bond.

burning is an example of exothermic reaction

atomic number of element is x the symbol of its ion is +1 bcz it will lose one electron

if temperature increases then rate of chemical reaction increases

Explanation:

i hope this will help you :)

7 0

3 years ago

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici

Brut [27]

Question:

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

Answer:

The mean activity coefficient for HCl solution is 0.78.

Explanation:

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

$E=E_{0}-0.0514 \mathrm{V} \ln \gamma$

As we know that $E_{0}$ = 0.22 V and E = 0.342 V, the equation will become

$0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma$

$0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)$

$0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)$

$\frac{0.12}{0.0514}=-\ln (0.124 \gamma)$

$-2.3346=\ln (0.124 \gamma)$

$e^{-2.3346}=0.124 \gamma$

$0.0968=0.124 \gamma$

$\gamma=\frac{0.0968}{0.124}=0.78$

So, the mean activity coefficient is 0.78.

6 0

3 years ago

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

3 0

2 years ago

What condition of the gas remains constant?

Alexeev081 [22]

Answer:

The temperature remains constant because the internal energy only depends on temperature in that case

-Hops

7 0

3 years ago

Changes to the global climate could include _______.

frosja888 [35]

1. Answer is d. all of the above.

Climate can be defined as the data collection of weather conditions of a long period of time in a certain area. The weather conditions that affect to the climate are temperature, precipitation, atmospheric pressure, wind and so on. Any changes in weather can change the global climate.

2. Answer is c. ozone.

Trees release oxygen and water vapor into the atmosphere but not ozone. The oxygen gas is released by trees as the by-product of photosynthesis. This is the natural way of production of oxygen gas. Water vapor can be released due to the transpiration of trees.

6 0

3 years ago

Read 2 more answers