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Novosadov [1.4K]
3 years ago
12

What part of cellular respiration is affected by arsenic

Chemistry
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

It effects all of the cellular respiration process

Explanation:

It inhibits the Glycolysis. It replaces the phosphate groups that is needed for making Pyruvate and ATP.

You might be interested in
5F248+ 2NH3(8) ► NF418 + 6HF
mylen [45]

Answer:

14.2L at STP

Explanation:

Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:

<em>Moles NH3 -Molar mass: 17.031g/mol-</em>

3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3

<em>Moles HF:</em>

0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF

<em>Volume HF</em>

PV = nRT; V = nRT/P

<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>

Replacing:

V = 0.634moles*0.082atmL/molK*273.15K / 1atm

V = 14.2L at STP

6 0
2 years ago
7. A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?​
USPshnik [31]

Answer:

The volume is

<h2>180 mL</h2>

Explanation:

In order to solve for the volume we use the formula for Boyle's law which is

<h3>P _{1}  V _{1} = P _{2}V _{2}</h3>

where

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Since we are finding the final volume we are finding V2

Making V2 the subject we have

<h3>V _{2}  = \frac{P _{1}  V _{1}}{P _{2}  }</h3>

From the question

P1 = 300 mmHg

V1 = 300 mL

P2 = 500 mmHg

Substitute the values into the above formula and solve for the final volume obtained

That's

<h3>V _{2} =  \frac{300 \times 300}{500}  \\  =  \frac{90000}{500}  \\  =  \frac{900}{5}</h3>

We have the final answer as

<h3>180 mL</h3>

Hope this helps you

7 0
3 years ago
Is water the only liquid? If not, name 5 more
guapka [62]

Answer:

Is water the only liquid? If not, name 5 more

Water, ethanol, household bleach, blood, paint, milk, gasoline, mineral oil, acetone and butyl alcohol are examples of liquids. Liquids' properties allow them to flow or be poured easily into containers, lava

water is not the only liquid.

6 0
3 years ago
Read 2 more answers
If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef
andrezito [222]

Answer:

B

Explanation:

Recall the law of effusion:

\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be <em>r</em>₁ and the rate of hydrogen be <em>r</em>₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for <em>r</em>₂:

\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\  r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}

Because there are 5 moles of hydrogen gas:

\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0
2 years ago
Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar=
IceJOKER [234]

Answer:

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

pH = -log[H⁺]

<h3>pH = 4.27</h3>

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

<h3>0.03%</h3>
5 0
2 years ago
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