Whichstatement describes a question that can be answered by a scientific investigation

Rus_ich [418]

Weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

7 0

2 years ago

Scientists at a food manufacturing company compare the properties of substances for use in flavoring frozen yogurt. They prepare

Ludmilka [50]

I pick but I'm not sure about it though 1and3

7 0

2 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

3 years ago

graduated cylinder is filled with water to a volume of 6.2 ML. an irregular shaped plastic object weighing 1.2 g is placed in th

Aleks04 [339]

Answer:

A. Density of object = 0.86 g/mL

B. The object will float in the water.

Explanation:

The following data were obtained from the question:

Volume of water = 6.2 mL

Mass (m) of object = 1.2 g

Volume of water + Object = 7.59 mL

Density of object =?

Density of water = 1 g/mL

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of water = 6.2 mL

Volume of water + Object = 7.59 mL

Volume of object =?

Volume of object = (Volume of water + Object) – (Volume of water)

Volume of object = 7.59 – 6.2

Volume of object = 1.39 mL

Therefore, the volume of the object is 1.39 mL

A. Determination of the density of the object.

Mass (m) of object = 1.2 g

Volume (V) of object = 1.39 mL

Density (D) of object =?

Density = mass /volume

Density = 1.2/1.39

Density of object = 0.86 g/mL

B. Determination of whether the object will float or sink.

Density of object = 0.86 g/mL

Density of water = 1 g/mL

From the above, we can see that the density of water is greater than that of the object. This implies that the object is lighter than water. Therefore, the object will float in the water.

8 0

3 years ago

How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Gnoma [55]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

4 0

3 years ago