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3 years ago

What fundamental building block makes up all matter in the universe

Chemistry

1 answer:

Sloan [31]3 years ago

6 0

The building block that makes up all matter in the universe is atoms.

Hope that helps you:)

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I need help with like 35 questions anyone willing to help please let me know I have discord

posledela

Answer and Explanation:

The molar mass of a substance is all of the weights from the elements combined.

So, we have the elements

CU, S, and $O_{4}$

CU has a mass is 65

S mass is 32

O has a mass of 16, but there's 4 atoms of O, so we do 16 times 4, which is 64.

Now we add.

65 + 32 + 64 is 161.

So, the answer is 160, or answer choice A.

#teamtrees #PAW (Plant And Water)

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3 years ago

Do any of you gies know how to bet Arceus in legend Arceus

diamong [38]

Best Team And Party in Pokemon Legends Arceus. Includes the best team for Cyndaquil, Oshawott, Rowlett, by type, early and middle game, where to catch, recommended Pokemon, & more!

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2 years ago

Claim: If I brush my teeth everyday, then I won't get cavities.

Rom4ik [11]

The dependent variable is i won’t get cavities

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4 years ago

Read 2 more answers

How many moles are in 1.806 x 1024 molecules of bromine?

Pepsi [2]

<h3>Answer:</h3>

2.999 mol Br

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

Step 1: Define

1.806 × 10²⁴ molecules Br

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$\displaystyle 1.806 \cdot 10^{24} \ molecules \ Br(\frac{1 \ mol \ Br}{6.022 \cdot 10^{23} \ molecules \ Br} )$ = 2.999 mol Br

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

Our final answer is already in 4 sig figs, so there is no need to round.

7 0

3 years ago

Read 2 more answers

A compound with molecular formula C6H15N exhibits a singlet at d 0.9 (1H), a triplet at d 1.10 (3H), a singlet at d1.15 (9H), an

LuckyWell [14K]

Answer:

N-ethyl-2-methylpropan-2-amine

Explanation:

In this case, we have to start with the IR info. The signal on 3400 cm^-1 indicates the presence of a hydrogen bonded to the heteroatom. In this case, we have nitrogen in the formula, so we will have the amine group.

On the other hand, we have to analyze the NMR info:

a) We have 2 singlets => This indicates the presence of 2 different hydrogens without neighbors.

b) We have a triplet => This indicates the presence of CH3 bonded to a CH2.

c) We have a quartet => This indicates the presence of CH2 bonded to a CH3.

From b) and c) we can conclude that we have the ethyl group bonded to a nitrogen.

Finally, we have to add 4 more carbons in such a way that we only have a single signal. In this case the ter-butyl group.

In that way, we will have 2 singlets (from the CH3 groups in the ter-butyl and the H on the N). Also, we will have the quartet on the CH2 in the ethyl group and the triplet on the CH3 in the ethyl group

6 0

3 years ago