What is the correct answer? Please

Describe step by step how we can make a field of a variable quantity?

just olya [345]

Finding the variables in the experiment or study and measurinf

7 0

3 years ago

HELP PLEASE

Andreas93 [3]

Answer:

A.)

Explanation:

The first structure accurately represents ethyl ethanoate because there is an ethyl group (2 carbons) directly attached to the oxygen within the parent chain and there are 2 carbons (the eth- in ethanoate) attached to the double bonded oxygen.

The second structure is not ethyl ethanoate. There is an ethyl group directly attached to the oxygen within the parent chain. However, there are 3 carbons attached to the double bonded oxygen. This makes the structure ethyl propanoate.

3 0

2 years ago

If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: 2KMnO4 + 5Hg2Cl2 + 16HCl -&gt

IceJOKER [234]

Answer:

The limiting reactant is Hg2Cl2.

Explanation:

Step 1: Data given

Mass of each reactant = 5.0 grams

KMnO4 MM=158 g/mol

Hg2Cl2 MM=472.1 g/mol

HCl MM=36.5 g/mol

HgCl2 MM=271.5 g/mol

MnCl2 MM=125.8 g/mol

KCl MM=74.6 g/mol

H2O MM=18 g/mol)

Step 2: The balanced equation

2KMnO4 + 5Hg2Cl2 + 16HCl -> 10HgCl2 + 2MnCl2 + 2KCl + 8H2O

Step 3: Calculate moles

KMnO4 = 5.00 grams / 158 g/mol = 0.0316 mol

Hg2CL2 = 5.00 grams / 472.1 g/mol = 0.0106 mol

HCl = 5.00 grams / 36.5 g/mol = 0.137 mol

Step 3: Calculate limiting reactant

For 2 moles of KMno4 we need 5 moles of Hg2Cl2 and 16 moles of HCl

Hg2Cl2 has the smallest amount of moles.

For 5 moles Hg2Cl2 ( 0.0106 mol) we need 0.0106 / (5/2) = 0.00424 mol KMnO4

For 5 moles Hg2Cl2 we need (16/5) *0.0106 = 0.03392 moles of HCl

So the limiting reactant is Hg2Cl2.

Step 4: Calculate moles of product produced:

2*0.0106 = 0.0212 moles of HgCl2

(2/5) * 0.0106 = 0.00424 moles of MnCl2 and 0.00424 moles of KCl

(8/5) * 0.0106 = 0.01696 moles H2O

7 0

3 years ago

In a cycle of copper experiment, a student first reacts a piece of copper metal with nitric acid to produce copper(II) nitrate (

LuckyWell [14K]

Answer:

0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment

Explanation:

The concentration of the solution is given by :

$[C]=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}$

We have:

Concentration of copper (II) nitrate solution = $[Cu(NO_3)_2]=2.41 M$

The volume of solution = 4.12 mL

1 mL= 0.001 L

$4.12 mL= 4.12\times 0.001 L= 0.00412 L$

Moles of copper (II) nitrate in solution = n

$2.41=\frac{n}{0.00412 L}=0.0099292 mol$

Moles of copper (II) nitrate in solution = 0.0099292 mol

1 Mole of copper(II) nitrate has 1 mole of copper then 0.0099292 moles of copper(II) nitrate will have :

$1\times 0.0099292 mol= 0.0099292 \text{ mol of Cu}$

Mass of 0.0099292 moles of copper:

$=0.0099292 mol\times 63.55 g/mol=0.63100 g\approx 0.631 g$

This mass of copper present in the solution is the theoretical mass of copper present in the given copper(II) nitrate solution.

0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment

7 0

3 years ago

What natural and human made features can a map show? Give two examples of each.

Bess [88]

Natural: Land mounds, Rivers
Human made: Country borders, cities

3 0

3 years ago