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4 years ago

The tendency for an object at rest to remain at rest is?

1 answer:

6 0

Newtons First Law... Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

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If 11.9 kJ are used to heat a sample of water the temperature increases from 20.0°C to

Kipish [7]

Answer:

$m=4.51g$

Explanation:

Hello!

In this case, since the energy involved during a heating process is shown below:

$Q=mCp\Delta T$

Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:

$m=\frac{Q}{Cp\Delta T}$

Thus, by plugging in, we obtain:

$m=\frac{11900J}{4.184\frac{J}{g\°C}(650\°C-20.0\°C)}\\\\m=4.51g$

Best regards!

7 0

3 years ago

How do you make a model of something you cannot see?

Vedmedyk [2.9K]

You have to imagine it

8 0

3 years ago

Read 2 more answers

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

3 years ago

Approximately 1 mL of two clear, colorless solutions-0.1 M Ba(NO3)2 and 0.1 M Na2SO4- were combined. Upon mixing, a thick, milky

murzikaleks [220]

Answer: balanced chemical equation: $Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Net ionic equation : $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation is:

$Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

$Ba^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2Na^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are sodium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

4 0

3 years ago

Which element has a larger atomic radius than sulfur? chlorine cadmium fluorine oxygen

zhuklara [117]

Answer:

Cadmium has larger atomic radius than sulfur.

Explanation:

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.

Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:

Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.

Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.

Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:

Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.

Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.

So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:

O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.

Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.

8 0

3 years ago