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3 years ago

Compare and contrast a pure substance with a mixture and give an example.

1 answer:

3 0

Of any substance? What about two powders such as 1/2 cup of white flour, then mixing 1/2 cup of salt together with the flour. This would then form a more coarse mixture being that the two particles differ in shape.

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Write the balanced molecular chemical equation for the reaction

Ierofanga [76]

The balanced molecular chemical equation for the reaction will be expressed as Cs₂CO₃ + Mg(NO₃)₂ -> 2CsNO₃ + MgCO₃

For any chemical equation to be balanced, the number of moles of elements in the reactants must be equal to that of the product.

According to the question, we are to write a balanced equation for the reaction in aqueous solution for cesium carbonate and magnesium nitrate

The chemical formula for Cesium carbonate is Cs₂CO₃

The chemical formula for magnesium nitrate is Mg(NO₃)₂

Hence the balanced molecular chemical equation for the reaction will be expressed as Cs₂CO₃ + Mg(NO₃)₂ -> 2CsNO₃ + MgCO₃

Learn more here: brainly.com/question/11904811

4 0

3 years ago

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James decides to walk home from school today. He lives 3 miles from school and can walk home in 45 minutes. At what rate is Jame

Natasha2012 [34]

One mile every 15 minutes

7 0

3 years ago

Please match the orbital type with the correct number of orbitals

saul85 [17]

Hi there!

p = e-3
s = f-1
f = i-7
d = g-5

Hope that helps!
Brady

7 0

3 years ago

Look at the following reaction. How could you increase the production of 2NH3(g)? N2(g) + 3H2(g) 2NH3(g) Increase the volume of

Zolol [24]

The correct option is B. To increase the production of ammonia, you have to increase the pressure of the system. Increase in pressure will result in increased production of ammonia because this will drive the chemical reaction forward.

5 0

3 years ago

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A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C

Mamont248 [21]

Answer:

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

6 0

3 years ago

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