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2 years ago

A lab experiment shows: merge a litmus paper in a

Chemistry

1 answer:

ddd [48]2 years ago

6 0

Answer:

Explanation:

When you put the paper in a solution, it will turn blue if it is basic, or red if it is acidic. If it does not change color, it is fairly neutral.

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How to do part (b)??

laila [671]

it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.

3 0

2 years ago

A student experimentally obtained the density of osmium the densest element as 22.57g/cm3 the density of osmium is reported to b

Airida [17]

Answer:

The answer to your question is: % error = 0.4

Explanation:

Data

real value = 22.48%

estimated value = 22.57 %

Formula

% error = |real value - estimated value|/real value x 100

%error = |22.48 - 22.57|/22.48 x 100

% error = |-0.09|/22.48 x 100

%error = 0.09/22.48 x 100

% error = 0.004 x 100

% error = 0.4

3 0

3 years ago

How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?

lesya692 [45]

Answer:

3.72 mol Hg

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Density = Mass over Volume

Explanation:

Step 1: Define

D = 13.6 g/mL

54.8 mL Hg

Step 2: Identify Conversions

Molar Mass of Hg - 200.59 g/mol

Step 3: Find

13.6 g/mL = x g / 54.8 mL

x = 745.28 g Hg

Step 4: Convert

$745.28 \ g \ Hg(\frac{1 \ mol \ Hg}{200.59 \ g \ Hg} )$ = 3.71544 mol Hg

Step 5: Check

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

3.71544 mol Hg ≈ 3.72 mol Hg

8 0

2 years ago

The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

Now put all the given values in this formula, we get:

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

3 years ago

PLSSS HELP. And there’s also at the bottom of the picture another option if u didn’t see it

WARRIOR [948]

Answer:

B. Gradualism

Explanation:

8 0

2 years ago