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2 years ago

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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7. A gas-filled weather balloon with a volume of 65.0 L is released at sea-level conditions of

Galina-37 [17]

The balloon will reach its maximum volume and it will burst.

Given:

A weather balloon at sea level, with gas at 65.0 L volume, 745 Torr pressure, and 25C temperature.
When the balloon was taken to an altitude at which temperature was 25C and pressure was 0.066atm its volume expanded.
The maximum volume of the weather balloon is 835 L.

To find:

Whether the weather balloon will reach its maximum volume or not.

Solution:

The pressure of the gas in the weather balloon at sea level = $P_1=745 torr$

$1 atm = 760 torr\\P_1=745 torr=\frac{745}{760} torr = 0.980 atm$

The volume of the weather balloon at sea level = $V_1=65.0L$

The temperature of the gas in the weather balloon at sea level:

$T_1=25^oC=25+273.15 K=298.15 K$

The balloon rises to an altitude.

The pressure of the gas in the weather balloon at the given altitude:

$P_2=0.066atm$

The volume of the weather balloon at the given altitude = $V_2=?$

The temperature of the gas in the weather balloon at the given altitude:

$T_1=25^oC=25+273.15 K=298.15 K$

Using the Combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\frac{0.980 atm\times 65.0L}{298.15 K}=\frac{0.066atm\times V_2}{298.15K}\\V_2=\frac{0.980 atm\times 65.0L\times 298.15K}{0.066atm\times 298.15 K}\\=965L$

The maximum volume of the weather balloon= V = 835 L

$V < V_2$

The volume of the weather balloon at a given altitude is greater than its maximum volume which means the balloon will reach its maximum volume and it will burst.

Learn more about the combined gas law:

brainly.com/question/13154969?referrer=searchResults

brainly.com/question/936103?referrer=searchResults

4 0

1 year ago

A single cell undergoes mitosis every five minutes. How many cells will result from this cell in 15 minutes? In 30 minutes?

Finger [1]

In 15 minutes 3 cells will be reproduced and in 30 minutes 6 cells will be reproduced

6 0

2 years ago

Read 2 more answers

A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is

musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁ = 1.08atm

V₁ = 250mL

T₁ = 24°C

P₂ = 2.25atm

T₂ = 37.2°C

V₂ = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

T₁ = 24°C, T₁ = 24 + 273 = 297K

T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K

Input the variables and solve for V₂

$\frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}$

V₂ = 124.91mL

6 0

2 years ago

Determine the number of moles and mass requested for each reaction in Exercise 4.42.

suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

Step 1: Write the balanced equation
Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. 46g of Na react with 1 mol of Cl₂. Since the molar mass of Cl₂ is 71g/mol, then 46g of Na react with 71g of Cl₂.

Step 3:

$10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}$

$10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}$

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

216.5g of HgO form 0.5 moles of O₂. 216.5g of HgO form 16g of O₂.

Step 3:

$1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}$

$1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}$

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

16g of O₂ come from 1 mol of NaNO₃. 16g of O₂ come from 85g of NaNO₃.

Step 3:

$128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}$

$128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}$

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

12 g of C form 1 mol of CO₂. 12 g of C form 44g of CO₂.

Step 3:

$20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2$

$[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2$ [/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

79.5g of CuO come from 1 mol of CuCO₃. 79.5g of CuO come from 123.5g of CuCO₃.

Step 3:

$1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}$

5 0

2 years ago

This is a document that each chemical manufacturer, distributor, or importer must provide for each hazardous chemical. It contai

kkurt [141]

Answer : HazCom

Explanation : Hazard communication which is also known as HazCom, is a set of processes and procedures that every employers and importers must implement in their workplace to effectively communicate hazards associated with chemicals during handling, shipping, and any form of exposure.

The OSHA Hazard Communication Standard is a U.S. regulation which governs the evaluation and communication of hazards associated with chemicals at the workplace. It is typically not attached to any specific chemical container but is stored in the workplace.

7 0

2 years ago

Read 2 more answers