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3 years ago

15

A submarine changes its mass by adding or removing seawater from tanks inside the sub. Explain how this can enable the sub to di

ve or rise to the surface.

1 answer:

bogdanovich [222]3 years ago

3 0

Less water = less weight to make it rise
More water = more weight to make it dive

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1 What characteristic of an element determines how its atoms can bond with other<br> atoms?

Masja [62]

Answer:

Electrons

Explanation:

The electrons of the outermost energy level determine the energetic stability of the atom and its tendency to form chemical bonds with other atoms to form molecules. Under standard conditions, atoms fill the inner shells first, often resulting in a variable number of electrons in the outermost shell.

8 0

4 years ago

Somebody please help me ?

statuscvo [17]

Its the second one. Add the exponents and you'll get 7

3 0

3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can <u>calculate the pH value</u>:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

Ernest Shackleton's South! primarily uses which writing structure?

damaskus [11]

Ernest Shackleton's South! primarily uses the writing structure of "problem and solution".

7 0

3 years ago

10. A 2.36-gram sample of NaHCO3 was completely decomposed in an

Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

$m_r=m_{NaHCO_3}=2.36 g$

We have two products formed, sodium carbonate and carbonic acid. This implies that the total mass of the products is:

$m_p=m_{Na_2CO_3}+m_{H_2CO_3}$

Apply the law of mass conservation:

$m_r=m_p$

Substitute the given variables:

$m_{NaHCO_3}=m_{Na_2CO_3}+m_{H_2CO_3}$

Rearrange for the mass of carbonic acid:

$m_{H_2CO_3}=m_{NaHCO_3}-m_{Na_2CO_3}=2.36 g - 1.57 g=0.79 g$

8 0

4 years ago