All categories

3 years ago

Complete the table by classifying each property as either a physical or chemical property

Chemistry

1 answer:

Advocard [28]3 years ago

8 0

Where is the table? I dont know what to classify

You might be interested in

Which of the following statements is not correct?

Lilit [14]

Answer:

It's 4.1 moles Li2O = 94 g on Odyssey ware

8 0

3 years ago

Read 2 more answers

I NEED HELP ! ANYTHING WILL HELP!

raketka [301]

Trilobites are more common than Sea urchins

5 0

3 years ago

PLZ HELP IM CONFUSED

Marina CMI [18]

Answer:

9.1 mol

Explanation:

The balanced chemical equation of the reaction is:

CO (g) + 2H2 (g) → CH3OH (l)

According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).

To convert 36.7 g of hydrogen gas to moles, we use the formula;

mole = mass/molar mass

Molar mass of H2 = 2.02g/mol

mole = 36.7/2.02

mole = 18.17mol

This means that if;

2 moles of H2 reacts to produce 1 mole of CH3OH

18.17mol of H2 will react to produce;

18.17 × 1 / 2

= 18.17/2

= 9.085

Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).

6 0

3 years ago

Number of electrons for all of these even the ones with a number by them ? Will give brainlist!!!!

Bumek [7]

Explanation:

s=2

p=6

d=10

f=14

pls mark me brainlist

7 0

2 years ago

The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i

Rus_ich [418]

Answer: a) The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$

7 0

3 years ago