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2 years ago

9

Magsulat sa iyong sailing lyrics na kanta the tune of meron leron sinta

1 answer:

xxMikexx [17]2 years ago

3 0

Answer:

I am sailing

I am sailing

Home again

'Cross the sea

I am sailing

Stormy waters

To be near you

To be free

I am flying

I am flying

Like a bird

'Cross the sky

I am flying

Passing high clouds

To be with you

To be free

Can you hear me, can you hear me

Through the dark night, far away

I am dying, forever crying

To be with you, who can say

Can you hear me, can you hear me

Through the dark night far away

I am dying, forever crying

To be with you, who can say

We are sailing, we are sailing

Home again

'Cross the sea

We are sailing

Stormy waters

To be near you

To be free

Oh Lord, to be near you, to be free

Oh Lord, to be near you, to be free

Oh Lord, to be near you, to be free

Oh Lord

Explanation:

I hope this is the right one

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Suppose you want to find out which dog food your dog likes best. Write about the experiment you would design. What variables do

denis-greek [22]

By giving the dog 2 or more different dog foods and seeing what one he eats the must of and the variables u control is how much you give the dog and the hypothesis is if I give the dog two or more different kinds of dog food than we will see what one he/she like the most

3 0

3 years ago

From the lists of available reagents select the one(s) you would use to in a preparation of acetophenone (phenyl methyl ketone)

VARVARA [1.3K]

Answer:

Step 1) hydrolysis using NaOH/H2O to form benzylalcohol

Step2) oxidation to Carboxylic acid using KMnO4 followed by decarboxylation to form benzene

3) friedel craft acylation using CH3COCl/AlCl3

Explanation:

The above 3 steps will yield acetophenone from methylbenzoate

3 0

2 years ago

Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s

NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

$n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3$

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0

2 years ago

According to the oxygen-hemoglobin dissociation curve, PO2 in the lungs of 100 mm Hg results in Hb being 98% saturated. At high

Ad libitum [116K]

Answer:

Hb would be 78.4% saturated.

Explanation:

This problem can be solved by using simple unitary method.

At 100 mm Hg pressure of oxygen, Hb is saturated by 98%

So, at 1 mm Hg pressure of oxygen, Hb is saturated by $\frac{98}{100}$ %

Hence, at 80 mm Hg pressure of oxygen, Hb is saturated by $\frac{98\times 80}{100}$ % or 78.4%

Therefore, at 80 mm Hg pressure of oxygen in the lungs, Hb would be 78.4% saturated.

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2 years ago

(a) In a gaseous sample of oxygen atoms and diatomic oxygen in equilibrium at 298K andunder a pressure of 1atm, what fraction of

ZanzabumX [31]

Answer:

Explanation:see below as shown from pics

7 0

3 years ago