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Veronika [31]
3 years ago
15

An autoclave is used to sterilize surgical equipment because

Chemistry
2 answers:
il63 [147K]3 years ago
8 0

Answer:

The correct option is D) it allows water to boil at temperatures above 100 ° C.

Explanation:

Hello!

Let's solve this!

In an autoclave, saturated water vapor is used, at a temperature of 121 ° C, at high pressures. In this way, it can be used to sterilize surgical and laboratory items.

The correct option is D) it allows water to boil at temperatures above 100 ° C.

avanturin [10]3 years ago
5 0

The answer to your question is D!

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Answer:

number

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A student isolated 15.6 g of product from a chemical reaction. She calculated that the reactions should have produced 18.4 g of
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Answer:

The percent yield of this reaction is 84.8 % (option A is correct)

Explanation:

Step 1: Data given

The student isolated 15.6 grams of the product = the actual yield

She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams

Step 2: Calculate the percent yield

Percent yield = (actual yield / theoretical yield ) * 100 %

Percent yield = (15.6 grams / 18.4 grams ) * 100 %

Percent yield  = 84.8 %

The percent yield of this reaction is 84.8 % (option A is correct)

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4 years ago
The majority of the elements essential to life are found in what part of the periodic table?
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3 years ago
How many oxygen atoms in 1 mole of CO2
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There are two oxygen atoms
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Consider the following reaction between mercury(II) chloride and oxalate ion:
siniylev [52]

Answer : The reaction rate will be, 1.9\times 10^{-4}M/s

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_2^{4-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b ....(4)

Dividing 1 from 2, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{k(0.164)^a(0.45)^b}{k(0.164)^a(0.15)^b}\\\\9=3^b\\(3)^2=3^b\\b=2

Dividing 3 from 2, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{k(0.164)^a(0.45)^b}{k(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2

Now, calculating the value of 'k' by using any expression.

Putting values in above rate law, we get:

3.2\times 10^{-5}=k(0.164)^1(0.15)^2

k=8.7\times 10^{-3}M^{-2}s^{-1}

Now we have to determine the reaction rate when the concentration of HgCl_2 is 0.135 M and that of C_2O_2^{-4} is 0.40 M.

\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2

\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2

\text{Rate}=1.9\times 10^{-4}M/s

Therefore, the reaction rate will be, 1.9\times 10^{-4}M/s

6 0
3 years ago
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