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3 years ago

The higher the altitude the colder the temperature? true or false.

Chemistry

2 answers:

hoa [83]3 years ago

5 0

It's True I Just Took The Test

seropon [69]3 years ago

4 0

The reason the temperature is colder is high altitude. It's colder at high altitude only if heat rises. The pressure in Earth's atmosphere, decreases with altitude.

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A liquid was analyzed to be 54.5% c, 9.10% h, and 36.4% o. an empty flask, whose mass was 45.32 g, when filled with the vapor of

nataly862011 [7]

<span>number Moles of C = 54.5 g / 12.011 = 4.54 number Moles of H = 9.10 / 1.008 = 9.02 number Moles of O = 36.4 / 16 = 2.28 if we want to divide by the smallest number 4.54 / 2.28 = 2 => C 9.02 / 2.28 = 4 => H 2.28 / 2.28 = 1 => O Empirical formula will be = C2H4O</span>

7 0

4 years ago

A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula

Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

8 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

According to the following reaction, how many grams of hydrogen

Mila [183]

Answer:

7.03 g

Explanation:

Step 1: Write the balanced synthesis reaction

N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)

Step 2: Calculate the moles corresponding to 32.5 g of N₂

The molar mass of N₂ is 28.01 g/mol.

32.5 g × 1 mol/28.01 g = 1.16 mol

Step 3: Calculate the number of moles of H₂ needed to react with 1.16 moles of N₂

The molar ratio of N₂ to H₂ is 1:3. The moles of H₂ needed are 3/1 × 1.16 mol = 3.48 mol.

Step 4: Calculate the mass corresponding to 3.48 moles of H₂

The molar mass of H₂ is 2.02 g/mol.

3.48 mol × 2.02 g/mol = 7.03 g

8 0

3 years ago

What is the reducing agent in the reaction below?

frez [133]

Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In $CrO_{3}$ , Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.

Hence, Cr is the oxidizing agent and Al is the reducing agent.

4 0

3 years ago

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