All categories

2 years ago

Significant Figures/Order of Operation Workshop

Chemistry

1 answer:

xz_007 [3.2K]2 years ago

4 0

Significant figures are as follows:

zero before (0.12 = 3 sig fig)

zero trapped (1.02 = 3 sig figs, or 0.012 = 3 sig figs)

zero on the end for whole number (120 = 2 sig figs)

zero on the end for decimal (1.20 = 3 sig figs)

Here are the first couple of answers:

1. 3

7. 2

13. 2

2. 2

8. 4

You might be interested in

Draw the product obtained when trans-2-butene is treated first with br2 in ch2cl2, second with nanh2 in nh3, and then finally wi

oee [108]

In order to do this, you have change the alkene into an alkyne. That is the aim of Br2/CH2Cl2 trailed by NaNH2. The Br2 with form a vic dihalide (3,4-dibromo octane). Adding of NaNH2 will execute two E2 reactions. -NH2 will eliminate an H from carbons 3 and 4. This double elimination will make the alkyne. Then handling the alkyne with H2/Lindlar will form the cis alkene. The final product will be CIS-3-octene.

3 0

2 years ago

How many moles of KCl will be produced from 15.0 g of KClO3?

slega [8]

Answer:

0.12 mol KCl

Explanation:

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

15 g x mol

x g KCl = 15 g KClO3 x[ (1 mol KClO3)/ (122.5 g KClO3) ] x [(2 mol KCl)/ (2 mol KClO3)]

x g KCl = 0.12 mol KCl

7 0

2 years ago

Read 2 more answers

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

A chemist is working on a reaction represented by this chemical equation:

erastova [34]

Answer is: A. 1.81 mol.

Balanced chemical reaction: FeCl₂ + 2KOH → Fe(OH)₂ + 2KCl.

n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

n(Fe(OH)₂) = 3.62 mol ÷ 2.

n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

6 0

2 years ago

Read 2 more answers

Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1

pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0

2 years ago