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3 years ago

What are the most chemically stable elements (in terms of reactivity) on the periodic table?

1 answer:

7 0

<span>Fluorine is the most chemically reactive element on the periodic table, and there are no chemical substances that are capable of freeing it from its bonds.

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Solve for a, b, c, AND d<br> d<br> C<br> 84°<br> 930<br> 970<br> b

olga55 [171]

Answer:

a = 87

b = 83

c = 96

d = 86

5 0

3 years ago

Read 2 more answers

NOT A TEST AND I WILL GIVE BRAINLEIST DON'T ANSWER IF YOU DON'T KNOW

barxatty [35]

Answer:

In the nucleus of an atom ,there are protons & neutrons. Protons have charge of 1.6× 10^-19 C, while neutrons have 0C charges. Electrons orbit outside the atom. Their charge is - 1.6 ×10^-19C

Explanation:

5 0

2 years ago

A fuel tank holds 22.3 gallons of gasoline. If the density is 0.8206 g/mL, what is the mass in kilograms of gasoline in a full t

7nadin3 [17]

Answer:

$m=69.3kg$

Explanation:

Hello!

In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

$V=22.3gal*\frac{3.78541L}{1gal}=84.415L$

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

$m=0.8206g/L*84.415L\\\\m=69.3kg$

Best regards!

8 0

3 years ago

The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium

stellarik [79]

Answer:

0.0177 L of nitrogen will be produced

Explanation:

The decomposition reaction of sodium azide will be:

$2NaN_{3}(s)--->2Na(s)+3N_{2}(g)$

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas

The molecular weight of sodium azide = 65 g/mol

The mass of sodium azide used = 100 g

The moles of sodium azide used = $\frac{mass}{molarmass}=\frac{100}{65}=1.54mol$

so 1.54 moles of sodium azide will give = $\frac{3X1.54}{2}=2.31$ mol

the volume will be calculated using ideal gas equation

PV=nRT

Where

P = Pressure = 1.00 atm

V = ?

n = moles = 2.31 mol

R = 0.0821 L atm / mol K

T = 25 °C = 298.15 K

Volume = $\frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L$

3 0

3 years ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<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>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

2 years ago