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3 years ago

Im teaching myself chemistry and need a guide. anyone game?

1 answer:

7 0

Which grade would u like to begin with? I can probably help with grade 9 Intro to Chemistry and Grade 10 Naming elements and compounds. That's all

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A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 1100 torr. The partial pressures of nitrogen, argon,

geniusboy [140]

Answer:

Partial pressure Ne = 340 Torr

Option B

Explanation:

Gases contained in the vessel:

N₂, Ar, He, Ne

One of Dalton's law for gases determine this:

In a mixture of gases contained in a vessel, total pressure of the system must be the sum of partial pressure of each gas.

Total pressure = 1100 Torr

Let's replace:

Partial pressure N₂ + Partial pressure Ar + Partial pressure He + Partial pressure Ne = 1100 Torr

Partial pressure Ne = 1100 Torr - Partial pressure N₂ - Partial pressure Ar -Partial pressure He

Partial pressure Ne = 1100 Torr - 110 Torr - 250 Torr - 400 Torr

Partial pressure Ne = 340 Torr

8 0

3 years ago

A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sampl

Rzqust [24]

Answer:

0.193 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 0.250 L

Initial temperature (T1) = 27°C

Initial pressure (P1) = 0.850 atm

Final volume (V2) =?

Final temperature (T2) = 0°C

Final pressure (P2) = 1.00 atm

Step 2:

Conversion of celsius temperature to Kelvin temperature.

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 27°C = 27°C + 273 = 300K

Final temperature (T2) = 0°C = 0°C + 273 = 273K

Step 3:

Determination of the new volume of the sample of ammonia gas.

The new volume can be obtain by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

0.850x0.250/300 = 1xV2/273

Cross multiply to express in linear form

300 x V2 = 0.850x0.250x273

Divide both side by 300

V2 = (0.850x0.250x273) /300

V2 = 0.193 L

Therefore, the volume of the sample at 0 °C and 1.00 atm is 0.193 L

6 0

3 years ago

Estimate the percent ionic character of the co bond. (note that the electronegativity of c is 2.5 and that of o is 3.5.)

deff fn [24]

From the calcuation, the percent ionic character of the bond is 70%

<h3>What is percent ionic character?</h3>

The term percent ionic character has to do with the degree of ionic bonding that is contained in a compound. It can be estimated from the electronegativity of each element.

We can use the formula; 100(1 - e^(-ΔEN² / 4))

EN = χB − χA * 100/1

EN = 3.5 - 1.5 = 1

100(1 - e^(-1)^2/4)

= 70%

Learn more about percent ionic character:brainly.com/question/7034446

#SPJ1

5 0

2 years ago

When 50 ml of 1.5 M HCl is added to 100 ml of 1.5 M sodium hydroxide solution, temperature

Georgia [21]

Answer:

Q = -1.318 KJ

Explanation:

We will use the assumption that this solution acts like water and thus we will use the specific heat capacity of water and when converting from mL to g, we will use the conversion like we do for water.

We are told that 50 mL of HCl reacts with 100 mL of NaOH.

Thus total mass; m = 50 + 100 = 150 mL

Converting to grams gives 150 g since we have assumed that the solution behaves like water.

We are given;

Initial temperature; T_i = 21.3° C

Final temperature; T_f = 23.4° C

ΔT = 23.4 - 21.3

ΔT = 2.1°C

Formula for quantity of heat is;

Q = mcΔT

c is specific heat capacity.

We will use c = 4.184 J/g°C since the solution is assumed to behave like water.

Thus;

Q = -(150 × 2.1 × 4.184)

Q = -1317.96 J

Negative sign is used because temperature was raised and thus reaction is exothermic.

Approximation to KJ gives; -1.318 KJ

7 0

3 years ago

If your end product is 200.0 g KMnO4 how much KOH did you start with?

aniked [119]

Answer:

$m_{KOH}= 142.0gKOH$

Explanation:

Hello there!

In this case, according to the following chemical reaction we found on goo gle as it was not given:

$2MnO_2+4KOH+O_2\rightarrow 2KMnO4+2KOH+H_2$

Whereas we can see a 2:4 mole ratio of potassium permanganate product to potassium hydroxide reactant with molar masses of 158.03 g/mol and 54.11 g/mol respectively. In such a way, by developing the following stoichiometric setup, we obtain the mass of KOH to start with:

$m_{KOH}=200.0gKMnO_4*\frac{1molKMnO_4}{158.03gKMnO_4}*\frac{4molKOH}{2molKMnO_4} *\frac{56.11gKOH}{1molKOH}\\\\m_{KOH}= 142.0gKOH$

Best regards!

8 0

3 years ago