The conversion of glucose to lactic acid is called:

Chemistry

2 answers:

weqwewe [10]3 years ago

7 0

Answer:

Anaerobic glycolysis

Explanation:

is the transformation of glucose to lactate when limited amounts of oxygen (O2) are available.

Yuki888 [10]3 years ago

4 0

Answer: Anaerobic glycolysis!

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. Use the following reaction to determine how much of each product would be released if 42 000 kg (42 tonnes) of methyl isocyana

fenix001 [56]

The amount of 1,3-dimethyl urea produced would be 32,458 grams or 32.458 kg while that of carbon dioxide would be 16,214 grams of 16.214 kg

<h3>Stoichiometric problem</h3>

From the equation of the reaction, the mole ratio of methyl isocyanate with the products is 2:1 respectively.

Mole of 42,000 kg of methyl isocyanate = 42000/57 = 736.84 moles

Equivalent mole of 1,3-dimethyl urea = 736.84/2 =368.42 moles

Equivaent mole of carbon dioxide = 736.84/2 =368.42moles

Mass of 368.42 moles 1,3-dimethyl urea = 368.42 x 88.1 = 32,458 grams or 32.458 kg

Mass of 368.42 moles of carbon dioxide = 368.42 x 44.01 = 16,214 grams of 16.214 kg

More on stoichiometric problems can be found here: brainly.com/question/14465605

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5 0

2 years ago

To determine the concentration of 20.00 mL of an unknown solution of a monoprotic acid, it is titrated with 0.1093 M sodium hydr

Elanso [62]

Answer:

d) 0.1202 M

Explanation:

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

NaOH + HA → NaA + H₂O

The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:

22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol

The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.

The molar concentration of HA is:

2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M

7 0

3 years ago

Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient

romanna [79]

Answer:

Coefficient of $H^{+}(aq)$ is more than 4

Explanation:

Oxidation: $Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance charge: $Sn^{2+}(aq)-2e^{-}\rightarrow Sn^{4+}(aq)$ ......(1)

Reduction: $Cr_{2}O_{7}^{2-}(aq)\rightarrow Cr^{3+}(aq)$

Balance Cr: $Cr_{2}O_{7}^{2-}(aq)\rightarrow 2Cr^{3+}(aq)$
Balance O and H in acidic medium: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$
Balance charge: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$ .......(2)