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2 years ago

What is the length in inches of a 100-meter field?

Chemistry

1 answer:

erastova [34]2 years ago

6 0

Answer: 3937 inches

Explanation: The conversion factor of 1m = 39.37 inches

Solution:

100 m x 39.37 in / 1 m

= 3937 inches

Cancel out meters so the remaining unit is in inches

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How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?

solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

7 0

2 years ago

Write a balanced net ionic equation for the following reaction. BaCI2(aq)+H2SO4(aq) -> BaSO4(s)+HCI (aq)

Sever21 [200]

Ba²⁺ + 2Cl⁻ + 2H⁺ + SO₄²⁻ = BaSO₄ (precipitate) + 2H⁺ + 2Cl⁻
Ba²⁺ + SO₄²⁻ = BaSO₄

4 0

2 years ago

Read 2 more answers

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

2 years ago

How many grams (m) of glucose are in 165 ml of a 5.50% (m/v) glucose solution? express your answer numerically in grams?

Leya [2.2K]

<span> mass of glucose = 0.055 *165 = 9.075 g

vol of methyl alc = 0.185 * 1.87 = 0.346 L = 346 ml

% NaCl ( m/v ) = mass NaCl * 100/ vol of soln

or Vol of Soln = mass NaCl / % NaCl (m/v)

= 32.1 * 100 / 6 = 535 ml the total vol of soln</span>

6 0

3 years ago

One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(

yuradex [85]

Answer:

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.

Explanation:

$Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)$

The balance chemical equation is :

$Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)$

Mass of barium hydroxide octahydrate = 6.5 g

Moles of barium hydroxide octahydrate = $\frac{6.5 g}{315 g/mol}=0.020635 mol$

According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:

$\frac{2}{1}\times 0.020635 mol=0.04127 mol$

Mass of 0.04127 moles of ammonium thiocyanate;

$0.04127 mol\times 76 g/mol=3.136 g\approx 3.14 g$

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate

3 0

2 years ago