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4 years ago

In what ways might a company that uses a strong acid handle an acid spill on the factory floor

Chemistry

1 answer:

In-s [12.5K]4 years ago

6 0

Well they would definitely rip up the flooring and replace it

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An element can be identified as either a ________, __________, or a __________

motikmotik

Liquid or a solid :)

5 0

4 years ago

Read 2 more answers

What is the molecular weight of a monoprotic carboxylic acid if 11.20 mL of 0.9635 M sodium hydroxide is required to titrate 1.2

Vlada [557]

Answer:

116.1 a.m.u.

It is not likely that RCOOH is the pentanoic acid

Explanation:

Let's consider the generic neutralization between NaOH and a monoprotic carboxylic acid.

RCOOH(aq) + NaOH(aq) ⇒ RCOONa(aq) + H₂O(l)

The molar ratio of RCOOH to NaOH is 1:1. The moles of RCOOH are:

$11.20 \times 10^{-3}L.\frac{0.9635molNaOH}{1L}. \frac{1molRCOOH}{1molNaOH} =1.079 \times 10^{-2}molRCOOH$

The molar mass of RCOOH is:

$\frac{1.253g }{1.079 \times 10^{-2}mol } =116,1g/mol$

Thus, the molecular weight is 116.1 a.m.u.

Pentanoic acid has the formula C₅H₁₀O₂ with a molecular weight of 102.1 a.m.u. So, it is not likely that RCOOH is the pentanoic acid.

4 0

4 years ago

How do you write a polynomial in standard form?

olchik [2.2K]

One way to write a polynomial is in standard form. In order to write any polynomial in standard form, you look at the degree of each term. You then write each term in order of degree, from highest to lowest, left to write. First, look at the degrees for each term in the expression. and in this way you will be able to write polynominal in standard form.

8 0

3 years ago

A dehydration reaction is the process in which __________.

Firlakuza [10]

Answer:

Dehydration reactions can be defined as the chemical reactions in which a water molecule is eliminated from the reactant molecule. The process of combination of two molecules with the elimination of water molecule is called dehydration synthesis.

3 0

2 years ago

Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you ne

katrin [286]

Answer:

7.08

Explanation:

To solve this problem we'll use the Henderson-Hasselbach equation:

pH = pka + log $\frac{[A^-]}{[HA]}$

Where $\frac{[A^-]}{[HA]}$ is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

pka = -log (1.8x10⁻⁴) = 3.74

We input the data:

4.59 = 3.74 + log $\frac{[A^-]}{[HA]}$

And solve for $\frac{[A^-]}{[HA]}$ :

0.85 = log $\frac{[A^-]}{[HA]}$

$10^{(0.85)}$ = $\frac{[A^-]}{[HA]}$

$\frac{[A^-]}{[HA]}$ = 7.08

3 0

3 years ago