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3 years ago

Convert 44 centimeters to meters. A. 4 m B. 4.4 m C. 0.44 m D. 0.044 m

Chemistry

2 answers:

Hoochie [10]3 years ago

8 0

The answer is C. 0.44 m

Sveta_85 [38]3 years ago

6 0

0.44 m is the right answer

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In a position time graph, a steep slope indicates _____.

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3 years ago

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What does knowing the temperature of a substance allow you to describe

il63 [147K]

Answer:

Ur answer is C. Their average kinetic energy

Explanation:

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2 years ago

If a laser light is compose of photons each with an energy of 4.58 × 10−19 J, what is the photon energy of this light in units o

kap26 [50]

Answer : The photon energy of this light in units of nanometers, (nm) is, $4.34\times 10^{-2}nm$

Solution :

Formula used :

$E=\frac{h\times c}{\lambda}$

where,

E = energy of photon = $4.58\times 10^{-19}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = ?

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

$4.58\times 10^{-19}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}$

$\lambda=4.3017\times 10^{-7}m=434.017\times 10^{-9}m=434.017nm=4.34\times 10^{-2}nm$

conversion used : $(1nm=1\times 10^{-9}m)$

Therefore, the photon energy of this light in units of nanometers, (nm) is, $4.34\times 10^{-2}nm$

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2 years ago

True or false ? Someone help me

zlopas [31]

Answer:

Yeah its true.................

Explanation:

True

3 0

2 years ago

How many atoms of aluminium are contained in 2.75 mol

horsena [70]

<h3>Answer:</h3>

1.66 × 10²⁴ atoms Al

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 2.75 mol Al

[Solve] atoms Al

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 2.75 \ mol \ Al(\frac{6.022 \cdot 10^{23} \ atoms \ Al}{1 \ mol \ Al})$
[DA] Multiply [Cancel out units]: $\displaystyle 1.65605 \cdot 10^{24} \ atoms \ Al$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.65605 × 10²⁴ atoms Al ≈ 1.66 × 10²⁴ atoms Al

4 0

2 years ago