In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:
Heat of the combustion reaction = mC(T2-T1)
= 1 (1.50) (41-21)
= 30 kJ
Answer:
it is probably acrylic or latex
Explanation:
Answer:
Positive
Explanation:
The antibody screen interpretation is positive. That's because of clump formation when antibody reacts with the antigen and agglutination occurs.
It means that the antibodies were present in the sample that react with the antigen present in the test tube.
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.

where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

The oxidation state of Pb in
is 2. So, moles deposited by Pb is as follows.

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
In reaction 1 of the Krebs cycle, acetyl‑CoA formed in the pyruvate dehydrogenase reaction condenses with the four‑carbon compound to form <em>citrate </em>with the elimination of coenzyme A. Since the product has three carboxyl groups, this pathway is referred to as the cycle. In reaction 2 of the Krebs cycle, this product then undergoes to form<em> isocitrate. </em>The enzyme is called aconitase because the compound cis‑aconitate is the <em>intermediate product</em> of the reaction. Reaction 3 eliminates CO2 to form the five‑carbon dicarboxylic acid <em>α-cetoglutarate. </em>Oxidation also occurs, with electrons transferred from the substrate to <em>COO-</em> . Consequently, this reaction is an oxidative decarboxylation.
In the image, you can see the reaction 2 in Krebs cycle is a two steps reaction with an intermediate cis-aconitase and a product called isocitrate.