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2 years ago

I can’t find the answer, can you help me please?

Chemistry

1 answer:

Tatiana [17]2 years ago

7 0

Answer: Warm ocean air then rises into the storm, then forming an area of low pressure underneath. :)

Explanation:

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Draw the diazonium cation formed when cytosine reacts with nano2 in the presence of hcl.

Sindrei [870]

Answer in the Word document below.
Diazonium compounds are a group of organic compounds sharing a common functional group R−N₂⁺. The process of forming diazonium compounds is called diazotation and usually <span>are prepared by treatment of aromatic amines with </span>nitrous acid<span> and additional acid (hydrochloric acid).
</span>Cytosine is one of the four main bases found in DNA and RNA.

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7 0

3 years ago

When 35.0 mL of 0.400 M hydrobromic acid and 35.0 mL of 0.400 M ammonia are combined, the pH of the resulting solution will be :

Kay [80]

Answer:

B. Equal to 7.

Explanation:

Hydrobromic acid is a strong acid that decreases pH and ammonia is a strong base that increases pH.

As the initial pH of water is 7,0 the addition of 35.0mL of 0.400M HBr will produce a pH less than 7,0. But, the same effect of decreasing pH is reverted for the addition of 35.0mL of 0.400M HNO3.

That means the net effect of the two addition is to have a pH:

B. Equal to 7.

I hope it helps!

3 0

2 years ago

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What

Gemiola [76]

Answer:

The answer to your question: 0.7 M

Explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

= 31.8

2 moles of KOH -------------- 1 mol of H₂SO₄

x -------------- 31.8 mol of H₂SO₄

x = (31.8)(2) / 1

x = 63.8 moles of KOH

Molarity = 63.8 / 90

= 0.7 M

7 0

2 years ago

Draw the best Lewis structure for CH 3 -1. What is the formal charge on the C?

djyliett [7]

Answer : The formal charge on the C is, (-1) charge.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $CH_3^{-1}$

As we know that carbon has '4' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in $CH_3^{-1}$ = 4 + 3(1) + 1 = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge on carbon atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on C}=4-2-\frac{6}{2}=-1$

The formal charge on the C is, (-1) charge.

3 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago