1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tems11 [23]
3 years ago
6

HCHllC CC001CH H2)C C1NO ,2BY​

Chemistry
2 answers:
Sever21 [200]3 years ago
8 0

Answer:

hakdog

Explanation:

pangitag lainbbbsjsbshsisbsjsbshxx

Alina [70]3 years ago
6 0
What does this mean man
You might be interested in
Pb(SO4)2 + 4 LiNO3 → Pb(NO3)4 + 2 Li2SO4
Anvisha [2.4K]

Answer:

4.5 moles of lithium sulfate are produced.

Explanation:

Given data:

Number of moles of lead sulfate = 2.25 mol

Number of moles of lithium nitrate = 9.62 mol

Number of moles of lithium sulfate = ?

Solution:

Chemical equation:

Pb(SO₄)₂ + 4LiNO₃      →     Pb(NO₃)₄ + 2Li₂SO₄

Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.

                       Pb(SO₄)₂        :         Li₂SO₄

                            1                :             2

                          2.25           :          2/1×2.25 = 4.5 mol

                       LiNO₃            :             Li₂SO₄

                           4                :                2

                           9.62           :             2/4×9.62 = 4.81 mol

Pb(SO₄)₂  produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of  Li₂SO₄.      

7 0
2 years ago
1. 90.0 mL of distilled water is added to a 10.0mL sample of 0.150mol/L sodium
aniked [119]

Answer: dilute

Explanation:

A concentrated solution which is used to prepare solutions of lower concentrations by diluting it with addition of water.

A dilute solution is one which contains lower concentration.

Using Molarity equation:

M_1 =concentration of stock solution = 0.150 mol/L

V_1 = volume of stock solution = 10.0 ml

M_2 = concentration of dilute solution = ?

V_2 = volume of dilute solution = (10.0+90.0) ml = 100.0 ml

0.150\times 10.0=M_2\times 100.0

M_2=0.015mol/L

As the concentration is less than the original concentration, the solution is termed as dilute.

7 0
3 years ago
What impact can limiting factors have on a population?
Salsk061 [2.6K]

Answer:

The answer is B. Limiting factors can lower birth rates, increase death rates.

7 0
2 years ago
Read 2 more answers
Current is described as
mrs_skeptik [129]
The answer is B the flow of electrons through a substance
3 0
3 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
3 years ago
Other questions:
  • Some elements are solids at room temperature true or false
    12·1 answer
  • The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
    11·1 answer
  • What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
    10·2 answers
  • C-12 has 6 protons. how many neutrons does c-13 have? select one:
    7·1 answer
  • The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
    8·1 answer
  • How many moles are in 2.3 grams of phosphorus
    12·1 answer
  • write a balanced net ionic equation for the following reaction: BaCl2(aq) + H2SO4 (aq) -- BaSO4(s) + HCl (aq)
    12·2 answers
  • Help me name these molecules please
    14·1 answer
  • how many grams of oxygen reacted with 50.7f of potassium? how many molecules of potassium oxide are formed?​
    9·2 answers
  • Describe how the periodic
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!