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3 years ago

If the mass of a certain block is 500g, what is the weight of the block?

1 answer:

5 0

______________________________

Solution,

Mass=500g =500/1000=0.5 kg

gravity(g)=9.8m/s^2

Now,

Weight= m*g

 = 0.5*9.8

 =4.9 N

So the weight of the block is 4.9 Newton

hope it helps...

Good luck on your assignment

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5000kg of ammonium nitrate per square kilometer of cornfield per year. how much nitric acid would be needed to make the fertiliz

Serggg [28]

$NH_3+HNO_3-> NH_4NO_3$

1 mole of nitric acid produce 1 mole of ammonium nitrate.

moles in 5000 kg of ammonium nitrate :

$n=\dfrac{5000\times 1000}{80}\\\\n=62500\ moles$ ( molecular mass of ammonium nitrate is 80 gm/mol )

So, number of moles of nitric acid required are also 62500 moles.

Mass of 62500 moles of nitric acid :

$mass = 62500\times 63 \\\\mass = 3937500\ gram\\\\mass = 393.75\ kg$

Hence, this is the required solution.

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2 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

CO_2 sublimes readily at 25°C. Which properties are usually associated with a compound that undergoes this kind of change?

antoniya [11.8K]

Explanation:

Sublimation is defined as a process in which solid state of a substance directly changes into vapor or gaseous state without undergoing liquid phase.

For example, naphthalene balls show sublimation at room temperature.

As this process does not cause any change in chemical composition of a substance. Hence, it is known as a physical process.

Similarly, when $CO_{2}$ sublimes readily at $25^{o}C$ . This shows change in physical state of carbon dioxide is taking place, i.e, from solid to gaseous phase.

Thus, we can conclude that when $CO_{2}$ sublimes readily at $25^{o}C$ then it means physical properties are usually associated with a compound that undergoes this kind of change.

4 0

2 years ago

1, 2, or 3...help please

Mashcka [7]

Answer:

molecular so number 3. ...

7 0

2 years ago

Read 2 more answers

Good day pleases help

klemol [59]

I can’t see the choices can you take another picture of this assignment please so i can help you