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3 years ago

HELP ASAP PLEASE I WILL MARK BRAINIST

Chemistry

2 answers:

Oduvanchick [21]3 years ago

6 0

Answer: The Answer Is C

Explanation:

jeka57 [31]3 years ago

4 0

It is iodine! I. Last answer.

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Consider the reaction below. 2Al2O3 4Al + 3O2 What is the mole ratio of oxygen to aluminum?

DiKsa [7]

The answer is 3/4.

The coefficient next to each reactant represents the amount of moles. The compound for oxygen is O2 and the compound for aluminum is 4. We can see that the number next to O2 is 3 and the number next to aluminum is 4.

8 0

3 years ago

Read 2 more answers

Please help me I would really appreciate it

denis-greek [22]

The number of students is your independent variable

6 0

3 years ago

In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o

Zigmanuir [339]

Answer:

Stock solution = 1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

Step 1: Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

Step 2: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

Step 3: Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

Step 4: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

Step 5: Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

Step 6: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0

3 years ago

If 5.0 g of copper metal reacts with a solution of silver nitrate, how many grams of silver metal are recovered?

sukhopar [10]

Answer:

16.9g

Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2

Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.

Then to calculate the theoretical yield you need to compare moles of the reactants:

m(Cu)=5g

M(Cu)=63.55

n(Cu)=5/63.55=0.0787

By comparing coefficients you require twice as much silver: 0.157mol

n(Ag)=0.157

M(Ag)=107.86

m(Ag)=0.157x107.86=16.9g

Hence, the theoretical yield of this reaction would be 16.9g

3 0

3 years ago

How many molecules of co2 in 97.3 grams of co2

s2008m [1.1K]

Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Reading a Periodic Table
Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

97.3 g CO₂

Step 2: Define conversions

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$97.3 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.33138 × 10²⁴ molecules CO₂

Step 4: Check

We are given 3 sig figs. Follow sig fig rules.

1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂

8 0

3 years ago