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2 years ago

Which statement is correct

Chemistry

2 answers:

alisha [4.7K]2 years ago

6 0

The answer to your question is A.

elixir [45]2 years ago

3 0

A is the correct answer

You might be interested in

Describe how you would prepare 350 ml of 0.100 m c12h22o11 starting with 3.00l of 1.50 m c12h22o11

Leona [35]

To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

Steps:

1. Take 23.33 mL of 1.50 M solution

<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M solution</span>

7 0

3 years ago

What’s the answer to this

Svetllana [295]

Answer:

B and D as well

Explanation:

All the electronic configuration of the elements of B end on s sub shell.

S energy level sub shell can hold 2 electrons and since Be, Mg, Ca Sr end in s.

We can say, they have same valence electrons, i.e 2 electrons.

6 0

2 years ago

Read 2 more answers

liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). I

max2010maxim [7]

The percent yield of carbon dioxide will be 49.0 %.

<h3>Percent yield</h3>

First, let's look at the equation of the reaction:

$2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O$

The mole ratio of octane to oxygen is 2:25.

Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol

Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol

Thus, octane is limiting.

Mole ratio of octane to carbon dioxide = 2:16.

Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol

Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams

Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %

More on percent yield can be found here: brainly.com/question/17042787

#SPJ1

6 0

1 year ago

Which of the following is a transuranium element?<br> Ra<br> Am<br> Tc<br> Pa

lyudmila [28]

Am - it has an atomic number of 95 which is greater than 92.

Transuranium elements are elements with atomic levels greater than 92

3 0

2 years ago

On the addition of HI

victus00 [196]

Answer:

$CH_3-CH(I)-CH(CH_3)-CH_3$

$I-CH_2-CH_2-CH(CH_3)-CH_3$

Explanation:

Given the compound $CH_2=CH-CH(CH_3)-CH_3$

The steps of reaction:

$HI \longrightarrow H^+ + I^-$

$CH_2=CH-CH(CH_3)-CH_3 + H^+ \longrightarrow CH_3-CH^+-CH(CH_3)-CH_3$

<u>But this intermediate product has a resonance: </u>

$CH_3-CH^+-CH(CH_3)-CH_3 \longleftrightarrow C^+H_2-CH2-CH(CH_3)-CH_3$

The reaction with I-

$CH_3-CH^+-CH(CH_3)-CH_3 + I^- \longrightarrow CH_3-CH(I)-CH(CH_3)-CH_3$

$C^+H_2-CH2-CH(CH_3)-CH_3 + I^- \longrightarrow I-CH_2-CH_2-CH(CH_3)-CH_3$

8 0

3 years ago