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3 years ago

A plane flew for 2 hours at 467 mph (mile per hour)and 5 hours at 536 mph. How far did the plane fly miles?

Chemistry

1 answer:

Andreyy893 years ago

7 0

Answer:

the plane flew 3614 miles

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“Kids and grown-ups love it so, the happy world of HARIBO!” Who does not like the dark brown liquorice and its lovely taste?! Ta

KiRa [710]

Answer:

Any Questions?

So i can give you the answer

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2 years ago

Use this oxidation-reduction reaction to answer questions about half-reactions:

Advocard [28]

Answer:

N, O, R

Explanation:

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5 0

3 years ago

How many moles are in 7.2x10^15 atoms of Pb?

fiasKO [112]

<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

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

<u>Math</u>

<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>

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] 7.2 × 10¹⁵ atoms Pb

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

6 0

3 years ago

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

CaHeK987 [17]

Answer:

Fe(CN)₂, FeCO₃, Pb(CN)₄, Pb(CO₃)₂

Explanation:

Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.

1.) Fe(CN)₂
-----> Fe²⁺ and CN⁻

-----> +2 + (-1) + (-1) = 0

2.) FeCO₃

-----> Fe²⁺ and CO₃²⁻

-----> +2 + (-2) = 0

3.) Pb(CN)₄

-----> Pb⁴⁺ and CN⁻

-----> +4 + (-1) + (-1) + (-1) + (-1) = 0

4.) Pb(CO₃)₂

-----> Pb⁴⁺ and CO₃²⁻

-----> +4 +(-2) + (-2) = 0

5 0

2 years ago

Will ya'll help me, please

JulsSmile [24]

Answer:

Either A or D

But i think D

hope this helps =3

sorry if im wrong

7 0

3 years ago