Students were asked to determine if the two slides are taken from plants or cells taken from animals

The incomplete table below shows selected properties of compounds that have ionic, covalent, or metallic bonds.

pogonyaev

Answer:

is there an option for silver? if so, silver is the answer.

8 0

1 year ago

a student prepares a dilute solution of sodium hydroxidem, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.

Serjik [45]

Answer:

M = 0.3077 M

Explanation:

As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.

Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.

Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:

n₁ = n₂ (1)

This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.

Now, we also know that moles can be expressed like this:

n = M * V (2)

And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.

The reported molar mass of KHP is 204.22 g/mol, so the moles are:

n = 1.372 / 204.22 = 6.72x10⁻³ moles

Now, we will use expression (2) to get the concentration of the diluted base:

n = M * V

M = n / V

M = 6.72x10⁻³ / 0.02184

M NaOH = 0.3077 M

This is the concentration of the dilute solution of NaOH

3 0

2 years ago

2. What does the text list and describe?

Mariana [72]

Answer:

C. ways radiation is transferred into and through Earth's atmosphere

Explanation:

6 0

2 years ago

A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and thi

dezoksy [38]

Answer:

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

Explanation:

Step 1: Data given

Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

Volume of a 0.860 M barium nitrate solution (Ba(NO3)2 = 14.8 mL = 0.0148 L

The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

Step 2: The balanced equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2KNO3(aq)

Step 3: Calculate moles

Moles = volume * molarity

Moles K2SO4 = 0.022 L * 1.16 M

Moles K2SO4 = 0.02552 moles

Moles Ba(NO3)2 = 0.0148 L * 0.860 M

Moles Ba(NO3)2 = 0.012728 moles

Step 4: Calculate the limiting reactant

For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

Ba(NO3)2 is the limiting reactant. It will completely be consumed. (0.012728 moles) . K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

Step 5: Calculate moles BaSO4

‬For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

For 0.012728 moles Ba(NO3)2 we'll have 0.012728 moles BaSO4

Step 6: Calculate mass BaSO4

Mass BasO4 = moles BaSO4 * molar mass BaSO4

Mass BaSO4 = 0.012728 moles * 233.38 g/mol

Mass BaSO4 = 2.97 grams

Step 7: Calculate the percent yield

% yield = (actual yield / theoretical yield ) * 100 %

% yield = ( 2.57 grams / 2.97 grams ) * 100 %

% yield = 86.5 %

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

7 0

2 years ago

Mixtures can be separated by physical changes. Please select the best answer from the choices provided

Mazyrski [523]

Where are the choices? :/

4 0

2 years ago

Read 2 more answers