Described the test for carbon dioxide. Give the result of the test.

What are the Group 1A and Group 7A elements examples of?

Scrat [10]

Alkali metals = Group 1A
Halogens = Group 7A

6 0

4 years ago

Read 2 more answers

1.Complete the balanced neutralization equation for the reaction below:

Alexxx [7]

1. $H_{2}$ S $O_{4}$ + Sr(OH) $_{2}$ ⇒ SrSO4 + 2 H2O is the balanced reaction.

2. 0.034 liters of KI will be required completely react with 2.43 g of Cu(NO3)2.

3. 0.55 liters is the volume in L of a 0.724 M Nal solution contains0.405 mol of NaI.

4. 33.3 ml many mL of 0.300 M NaF would be required to make a 0.0400 M solution of NaF when diluted to 250.0 mL with water

Explanation:

Balance chemical reaction of neutralization:

1. $H_{2}$ S $O_{4}$ + Sr(OH) $_{2}$ ⇒ SrSO4 + 2 H2O

2. Data given:

balance chemical reaction:

2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + I₂(s) + 4KNO₃(aq)

molarity of KI = 0.209 M

mass of Cu(NO₃)₂ = 2.43 grams

number of moles of Cu(NO₃)₂ will be calculated as:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

atomic mass of Cu(NO₃)₂ = 187.56 grams/mole

putting the values in the equation,

number of moles= $\frac{2.43}{187.56}$

= 0.0129 moles

2 moles of Cu(NO₃)₂ will react with 4 moles of KI

0.0129 moles will react with x moles of KI

$\frac{4}{2}$ = $\frac{x}{0.0129}$

x = 0.0258

atomic mass of KI = 166.00 grams/mole

mass = 166.00 x 0.0258

= 4.28 grams or ml is the final volume of KI

so molarity = $\frac{number of moles}{volume in litres}$

so molarity of KI is 0.0258 M, volume is 1 litre.

Using the formula

Minitial x Vinitial = M final Vfinal

V initial = $\frac{M final Vfinal}{Minitial}$

= $\frac{4.28 X 0.209}{0.0258}$

= 34.67 ml 0.034 liters of KI will be required.

3) Data given:

molarity of NaI = 0.724

number of moles of NaI =?

Volume in litres =?

formula used:

molarity = $\frac{number of moles}{volume in liters}$

volume in litres = $\frac{0.405}{0.724}$

= 0.55 liters is the volume

4) Data given:

Initial molarity = 0.3 M

initial volume = ?

final molarity = 0.04 M

final volume diluted by = 250 ml

formula used:

M initial X Vinitial = Mfinal X V final

putting the values in the equation:

Vinitial = $\frac{0.04 X 250}{0.3}$

= 33.3 ml of 0.3 M solution will be required.

5 0

4 years ago

Why do the ocean currents move from the equator towards the poles?

elena-s [515]

Hey there!

The reason why <span>ocean currents move from the equator towards the poles are because the wind push the water's away from the equator and toward the poles, so pretty much, let's blame this on the wind, it's the wind's fault not ours. But yes, your correct answer to this question above would be of that the wind's push the water's away from the equator and to the poles.

Hope this helps you buddy!</span>

4 0

4 years ago

How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work

strojnjashka [21]

2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!

8 0

4 years ago

12 G of carbon react with 16 G of oxygen how much carbon monoxide is formed

elena55 [62]

Answer:

28 g CO

Explanation:

First convert grams to moles.

1 mole C = 12.011 g (I'm just going to round to 12 for the sake of this problem)

12 g C • $\frac{1 mol C}{12 g C}$ = 1 mol C

1 mol O = 15.996 g (I'm just going to round to 16)

16 g O • $\frac{1 mol O}{16 g O}$ = 1 mol O

So the unbalanced equation is:

$C + O_{2}$ -> $CO$ (the oxygen has a 2 subscript because it is part of HONClBrIF meaning when not in a compound these elements appear in pairs - called diatomic elements)

The balanced equation is:

$2 C + O_2$ -> $2 CO$

However, carbon is the limiting reactant in this equation and two moles cannot react because only 12 g (1 mole) are present. Therefore, use the equation

$C + \frac{1}{2} O_2$ -> $CO$ .

1 mole of CO is formed, therefore 12 g + 16 g = 28 g CO.

3 0

3 years ago