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tia_tia [17]
3 years ago
14

What is an example of force pair​

Chemistry
1 answer:
mina [271]3 years ago
3 0

he interaction between a baseball bat and a baseball

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How are sugar molecules in living things classified
AlekseyPX

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Inside the human body they are known as glucose.

Explanation:

Your pancreas secretes insuline through the body in order to mantain a proper glucose levels through out the body. This is also what causes diabetes. If someone has diabetes this means that the pancreas is having trouble mantaining glucose through the body. This is why you should not consume sugar very often, then your body forgets how to create its own supply of glucose

6 0
4 years ago
Read 2 more answers
odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add
Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, 6.54\times 10^3kg

Explanation :

First we have to calculate the moles of H_2SO_4.

\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}

Given:

Molar mass of H_2SO_4 = 98 g/mole

Mass of H_2SO_4 = 6.05\times 10^3kg=6.05\times 10^6g

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol

The moles of H_2SO_4 is, 6.17\times 10^4mol

Now we have to calculate the moles of Na_2CO_3

The balanced neutralization reaction is:

Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3

From the balanced chemical reaction we conclude that,

As, 1 mole of H_2SO_4 neutralizes 1 mole of Na_2CO_3

So, 6.17\times 10^4mol of H_2SO_4 neutralizes

Now we have to calculate the mass of Na_2CO_3

\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3

Molar mass of Na_2CO_3 = 106 g/mole

\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg

Thus, the mass of sodium carbonate added to neutralize must be, 6.54\times 10^3kg

3 0
4 years ago
If a balloon containing 3000 L of gas at 39°C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature
ludmilkaskok [199]

\tt =3000~L\times \dfrac{289}{312}\times \dfrac{99}{45.5}

<h3>Further explanation</h3>

Given

3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,

Required

the new volume

Solution

Combined with Boyle's law and Gay Lussac's law  

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

T₁ = 39 + 273 = 312

T₂ = 16 + 273 = 289

Input the value :

V₂ = (P₁V₁.T₂)/(P₂.T₁)

V₂ = (99 x 3000 x 289)/(45.5 x 312)

or we can write it as:

V₂ = 3000 L x (289/312) x (99/45.5)

3 0
3 years ago
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