It is normal to heal after having sex :c?

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marusya05 [52]

The answer should be d

6 0

3 years ago

In the sodium-potassium pump, sodium ion release causes two K ions to be released into the cell. __________________ True False

Solnce55 [7]

Answer:

false

Explanation:

As we know that in sodium-potassium pump .

sodium potassium move 3Na+ outside the cells

and moving 2k+ inside the cells

so that we can say that given statement is false

Answer FALSE

6 0

2 years ago

What is the percentage of oxygen in Li(NO2)3

BartSMP [9]

Answer:

66.2 % of O

Explanation:

Our compound is the lithium nitrite.

LiNO₂

This salt is ionic and can be dissociated: LiNO₂ → Li⁺ + NO₂⁻

We determine the molar mass:

molar mass of Li + 3 . molar mass of N + 6 . molar mass of O

6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol

The mass of oxygen contained in 1 mol of lithium nitrite is:

6 . 16 g/mol = 96 g

So the percentage of oxygen present is:

(96 g / 144.94 g) . 100 = 66.2 %

3 0

3 years ago

Using Equation (10), calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl- ion. (Ecell in Equation (10) is the ne

mixer [17]

Answer:

7.16x10⁻⁸M = [Ag+]

Explanation:

Using the equation:

E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)

Where E⁰= 0.4249V

E(Cell) = -(-0.0019V) -Measured value-

[Cu2+] = 1M

Replacing:

0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)

-0.423V = - 0.0296 • log (1M/[Ag+]²)

14.29 = log (1M/[Ag+]²)

1.95x10¹⁴ = 1M / [Ag+]²

[Ag+]² = 5.12x10⁻¹⁵M

7.16x10⁻⁸M = [Ag+]

5 0

2 years ago

if 0.40 mol of h2 and .15 mol of o2 were to reat as completely as possible to produce h20, what mass of the reactant would remai

Sveta_85 [38]

Answer:

0.2g

Explanation:

Given parameters:

Number of moles of H₂ = 0.4mol

Number of moles of O₂ = 0.15mol

Unknown:

Mass of reactant that would remain = ?

Solution:

To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.

The expression of the reaction is :

2H₂ + O₂ → 2H₂O

2 mole of H₂ will combine with 1 mole of O₂

But given; 0.4 mole of H₂ we will require $\frac{0.4}{2}$ = 0.2mole of O₂

The given number of oxygen gas is 0.15mole and it is the limiting reactant.

Hydrogen gas is in excess;

1 mole of oxygen gas will combine with 2 mole of hydrogen gas

0.15 mole of oxygen gas will require 0.15 x 2 = 0.3mole of hydrogen gas

Now, the excess mole of hydrogen gas = 0.4 mole - 0.3 mole = 0.1mole

Mass of hydrogen gas = number of mole x molar mass

Molar mass of hydrogen gas = 2(1) = 2g/mol

Mass of hydrogen gas = 0.1 x 2 = 0.2g

8 0

3 years ago