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3 years ago

Which of these represent an unbalanced equation with the correbt chemical formuals

1 answer:

7 0

i can't give you a full answer to the lack of info, but everything goes somewhere. take the making of water, for instance. 2 Hydrogen and 1 Oxygen <u>must</u> equal H2O (2 parts hydrogen and 1 part oxygen) after being made.

Long story short: what goes in must have the same amount of atoms after the chemical reaction.

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Its B i already got it

Rina8888 [55]

Explanation:

With all the scaling up and design improvements, the flight range of the R-7 increased to an intercontinental distance from 330 kilometers for the V-2. The R-7 ended up looking oversized even next to its American contemporaries -- the Atlas and Titan ballistic missiles.

4 0

2 years ago

What is more electronegative? <br> O or Br

cluponka [151]

Answer:

Explanation:

O = 3.44

Br = 2.96

8 0

2 years ago

Give the n and l values and the number of orbitals for sublevel 5g.

Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4 ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1) ( 2×4 + 1) = 9

Thus, The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

brainly.com/question/14650894

#SPJ1

6 0

1 year 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:

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

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

Ort

ivanzaharov [21]

Answer:

1 ethanol is right answer

Explanation:

CH3- CH2-OH

4 0

3 years ago

Read 2 more answers