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3 years ago

Draw the structure 2 butylbutane

Chemistry

1 answer:

k0ka [10]3 years ago

5 0

Answer:

please look at the picture below.

Explanation:

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What is the mass of 99.0 L of NO2 at STP

Mariana [72]

At STP one mole of any gas occupies 22.4 L

moles NO2 = 99.0/22.4 = 4.42
mass NO2 = 4.42 mol x 46.0 g/mol=203 g

3 0

3 years ago

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago

What is the mass of 2.23 × 10^23 atoms of aluminum?

rjkz [21]

Answer:

Explanation:

In image

Take atomic mass or molar mass

of Al =27

7 0

2 years ago

Enter the balanced NET IONIC equation for the potentially unbalanced equation AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq).

Zigmanuir [339]

Answer:

$Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$

Explanation:

Hello!

In this case, since the net ionic equations are ionic representations of the molecular equation in which the spectator ions (those at both reactants and products sides) are cancelled out, we first write the complete ionic equation for this reaction, considering that the solid silver chloride is not ionized due to its precipitation:

$Ag^+(aq)+NO_3^-(aq)+Na^+(aq)+Cl^-(aq)\rightarrow AgCl(s)+Na^+(aq)+NO_3^-(aq)$

Whereas the nitrate and sodium ions are cancelled out for the aforementioned reason as they are the spectator ions, to obtain:

$Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$

Which is the required net ionic equation.

Best regards!

4 0

2 years ago

A rectangular solid has the following measurements 6 cm, 10 cm, and 5 cm. what is the volume of the solid?

AysviL [449]

300cm cubed ( have a great night! )

7 0

2 years ago