All categories

3 years ago

True or false all cells are the same size

Chemistry

2 answers:

MariettaO [177]3 years ago

8 0

Answer:

False

Explanation:

Nuetrik [128]3 years ago

6 0

Answer:

false

Explanation:

yes

You might be interested in

What do you mean by environment conservation?

BARSIC [14]

Explanation:

Environment conservation is an umbrella term that defines anything we do to protect our planet and conserve its natural resources so that every living thing can have an improved quality of life. bolivianouft and 7 more users found this answer helpful.

3 0

3 years ago

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

Donde:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

5 0

3 years ago

How does government ensure a safe environment for inverters?

Anestetic [448]

hope it helps please brainliest

7 0

3 years ago

Predict the following product(s) of the following reaction? F2 + KCI →

slega [8]

Answer:

${ \rm{F _{2(g)} + 2KCl _{(aq)} → { \bold{2KF _{(g)} + Cl_{2(g)} }}}}$

Explanation:

» The prediction is 98% correct because single displacement reaction type is highly possible.

This is because Fluorine has is more electronegative than Chlorine in Potassium Chloride. So, it strongly displaces Chlorine from the solution hence forming Chlorine gas.

» The 2% of wrong prediction maybe because of wrong reactant measurements following mole concept chemistry.

If you are asked the observation,

Observation » A green yellowish gas is formed.

This gas is Chlorine gas (Cl2)

5 0

2 years ago

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co

DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: $\frac{289kg}{1249 kg}$ =0,231 kg O₂/ kg air

I hope it helps!

4 0

3 years ago