How much 3m stock solution would it take to make 70 ml of 1.2 m solution?

What is the total number of elements in group 17 that are gases at room temperature and standard pressure?

White raven [17]

Answer:

Only to elements are gases at room temperature and standard pressure.

Fluorine and chlorine.

Explanation:

The group 17 is called halogen.

There are five elements in halogen group.

All halogens required one electrons to complete the octet and to get the noble gas electronic configuration.

All halogen elements have seven valance electrons.

Halogen elements:

Fluorine, chlorine, Bromine, iodine and astatine

Fluorine is gas at room temperature which s 25°C and standard pressure which is 1 atm.

It is present in the from of F₂.

Chlorine is also gas at room temperature and standard pressure.

It is present in the from of Cl₂.

The bromine is liquid under these condition.

Iodine and astatine are solids.

As we move down the group their melting and boiling points increases.

4 0

3 years ago

When bacteria reproduces asexually are the offspring uniform or diverse?

mario62 [17]

Answer:

Reproduction may be asexual when one individual produces genetically identical offspring, or sexual when the genetic material from two individuals is combined to produce genetically diverse offspring.

4 0

3 years ago

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You conduct an experiment that requires the creation of an ammonia solution. You do this by reacting 50.0 L of nitrogen gas with

ozzi

Answer : The molarity of the resulting ammonia solution is, 0.89 M

Explanation :

The balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

First we have to calculate the moles of nitrogen gas.

As we know that at STP, 1 mole of gas occupies 22.4 L volume of gas.

As, 22.4 L volume of nitrogen gas present in 1 moles of nitrogen gas

So, 50.0 L volume of nitrogen gas present in $\frac{50.0}{22.4}=2.23$ moles of nitrogen gas

Thus, the moles of nitrogen gas is 2.23 moles.

Now we have to calculate the moles of ammonia gas.

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 2.23 moles of $N_2$ react to give $2.23\times 2=4.46$ moles of $NH_3$

Now we have to calculate the molarity of the resulting ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of ammonia}}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{4.46mole}{5.0L}$

$\text{Molarity}=0.89M$

Therefore, the molarity of the resulting ammonia solution is, 0.89 M

7 0

3 years ago

3 Aqueous copper(II) sulfate solution is placed in an

Mazyrski [523]

Answer:

it will option D green to blue because iron being more reactive displaces copper from its solution

6 0

3 years ago

10.00 mL of the final acid solution is reacted with excess barium chloride to produce a precipitate of barium sulfate (Fw: 233.4

Verizon [17]

Answer:- Actual molarity of the original sulfuric acid solution is 17.0M.

Solution:- Barium chloride reacts with sulfuric acid to make a precipitate of barium sulfate. The balanced equation is written as:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

From this equation there is 1:1 mol ratio between barium sulfate and sulfuric acid. So, if excess of barium chloride is added to sulfuric acid then the moles of sulfuric acid would be equivalent to the moles of barium sulfate. Moles of barium sulfate could be calculated from the mass of it's dry precipitate.

Molar mass of barium sulfate is 233.4 grams per mol. The calculations for the moles of sulfuric acid are given below:

$0.397gBaSO_4(\frac{1molBaSO_4}{233.4gBaSO_4})(\frac{1moH_2SO_4}{1molBaSO_4})$

= $0.00170molH_2SO_4$

From given information, 10.00 mL of final acid solution were taken to react with excess of barium chloride. It means 0.00170 moles of sulfuric acid are present in 10.0 mL of final acid solution. We could calculate the actual molarity of the final solution from here as:

10.0 mL = 0.0100 L

$molarity=\frac{0.00170mol}{0.0100L}$

= 0.170M

Now we would use the dilution equation to calculate the actual molarity of the original sulfuric acid solution. The molarity equation is:

$M_1V_1=M_2V_2$

From given information, 10.0 mL of original acid solution were taken in a 100 mL flask and water was added up to the mark. It means the 10 fold dilution is done. 10 fold dilution means the molarity becomes one tenth of it's original value. Let's do the calculations in reverse way as we have calculated the molarity of the final solution.

let's say the molarity after first dilution is Y. the volume is taken as 10.0 mL. Final volume is 100 mL and the molarity is 0.170M. Let's plug in the values in the equation:

Y(10.0mL) = 0.170M(100mL)

$Y=\frac{0.170M*100mL}{10.0mL}Y = 1.70MLet's do the similar calculations to find out the actual molarity of the original acid solution. Let's say the molarity of the original acid solution is X. 10.0 mL of it were taken and diluted to 100 mL on adding water. The molarity is 1.70M as is calculated in the above step. Let's plug in the values in the molarity equation again to solve it for X as:X(10.0mL) = 1.70M(100mL)[tex]X=\frac{1.70M*100mL}{10.0mL}$

X = 17.0M

Hence, the actual molarity of sulfuric acid solution is 17.0M.

5 0

3 years ago