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jeka94
2 years ago
12

A chemist added hydrochloric acid to a sample of baking soda in a beaker. Immediately, the mixture of the two chemicals began to

fizz and bubble. Which of the following observations about the fizzing mixture provides evidence of a chemical
reaction?

A. The temperature of the beaker did not change.


B. A substance escaped from the beaker in the form of a gas.


C. Some of the baking soda remained in the bottom of the beaker.


D. Most of the baking soda appeared to dissolve in the acid.
Chemistry
2 answers:
Temka [501]2 years ago
4 0

Answer: B

Explanation:

tester [92]2 years ago
3 0

Answer: B

Explanation:

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When 500 g of water is cooled from 80.0°C to 10.0°C, how much heat energy is lost?
vampirchik [111]

I don’t know

Explanation:

6 0
2 years ago
Read 2 more answers
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
I am in desperate need of help will give 40 points to whoever helps me!!!!! I have a quiz tomorrow but I have no idea how to do
serious [3.7K]
All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.<span>
Freezing point depression or Boiling point elevation:

</span><span>ΔT = -K (m) (i)

</span>ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent. 
<span>
K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent. 
</span><span>
m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent. 
</span><span>
i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.</span>
5 0
3 years ago
or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

4 0
3 years ago
Non-metal ions are negative because...
abruzzese [7]

Explanation:

its b cz the gain electrons i think

6 0
3 years ago
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