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Drupady [299]
2 years ago
13

What is the starting molecule for glycolysis?

Chemistry
1 answer:
meriva2 years ago
5 0
Glucose is the starting molecule for glycolysis.
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How many moles of O2 would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 73.3% yi
Airida [17]

Answer:

Mmmmmmmmmmmmmmm

Explanation:

4 0
1 year ago
A pharmacist–herbalist mixed 100 g lots o St. John’s wort containing the ollowing percentages o the active component hypericin:
Anna [14]

Answer:

strength of hypericin in mixture = 0.42 %

Explanation:

given data

each lot = 100 g

active component hypericin = 0.3%, 0.7%, and 0.25%

solution

we get here percent strength o hypericin in the mixture that is

Hypericin contribution lot 1 =  \frac{0.3}{100} × 100

Hypericin contribution lot 1 =  0.3 g

and

Hypericin contribution lot 2 = \frac{0.3}{100} × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = \frac{0.25}{100} × 100  

Hypericin contribution lot 3  = 0.25 g

so

total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g

total 300 g mixture of hypericin = 1.25 g  

so here percent strength o hypericin in mixture is

strength of hypericin in mixture = \frac{1.25}{300} × 100  

strength of hypericin in mixture = 0.42 %

5 0
1 year ago
How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu
Alexxandr [17]
The  moles  of  chromium (iii)  nitrate  produced  is  calculated  as   follows

write  the  equation  for  reaction

 3  Pb(NO3)2  +  2 Cr  =  2 Cr(NO3)3  +  3  Pb

by  use  of  mole  ratio  between  Pb(NO3)2  to  Cr(NO3)3  which  is  3  :  2  the  moles  of  Cr(NO3)3  is therefore  
=  0.85  x2  /3  =  0.57   moles
5 0
1 year ago
What is the molarity of 4.35 moles kmno4 dissolved in 750 ml of solution?
Andreyy89
M= #moles / L
 
4.35/.75 = 5.6
7 0
1 year ago
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
1 year ago
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