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mihalych1998 [28]
2 years ago
7

0. 01 M HCl solution has a pH of 2. Suppose that during the experiment, both the universal pH indicator and the cabbage indicato

r turn orange-red for 0. 01 M HCl. What can you conclude about the the cabbage indicator key? It matches the universal pH indicator and is indicating the proper pH. It should completely match the universal indicator key for all pH values greater than 7 (base). It should completely match the universal indicator key for all pH values. There was some sort of experimental error, because the indicators should never match.
Chemistry
2 answers:
frosja888 [35]2 years ago
6 0

Answer:

Explanation:

An object is thrown vertically from the ground upwards with a speed of 10 m/s. Considering g = 10 m/s2, the maximum height that the object reaches from the ground, in meters, will be:

a) 15.0.

b) 10.0.

c) 5.0.

d) 1.0.

e) 0.5.

8090 [49]2 years ago
5 0

Answer:

a.) It matches the universal pH indicator and is indicating the proper pH.

Explanation:

it says its the answer.

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Many homeowners treat their lawns with CaCO3(s) to reduce the acidity of the soil. Write a net ionic equation for the reaction o
Agata [3.3K]

The reaction of acid, assuming HCl and calcium carbonate always produces a gas. The reaction is as follows:
2 HCl + CaCO3 --> CaCl2 + H2CO3 
H2CO3, carbonic acid, is a weak acid that is unstable in water solutions at high concentrations. As such, it decomposes: 
H2CO3 --> H2O + CO2 
Then, 
2 HCl + CaCO3 --> CaCl2 + H2O + CO2 
The total ionic equation looks as follows: 
2H+(aq) + 2 Cl-(aq) + CaCO3(s) --> Ca+2(aq) + 2 Cl-(aq) + H2O(l) + CO2(g) 
Clearly, Cl- is a spectator ion as it is unchanged in the reaction. The net ionic reaction looks as follows: 
2 H+(aq) + CaCO3(s) --> Ca+2(aq) + H2O(l) + CO2(g)
4 0
3 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0
2 years ago
1. A solution is always a mixture (true or false)
lakkis [162]

Answer:

yes solution is always a mixture but not all mixtures are solution

Explanation:

A solution.is a homogeneous mixture of substance that have uniform composition throughout

And a mixture hVe twoo or more substances that are not chemically.combine

4 0
3 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
3 years ago
Oxygen has an electronegativity of 3.5, and carbon has an electronegativity of 2.5. how is charge distributed on an oxygen atom
vlabodo [156]
When oxygen has an electronegativity of 3.5, and carbon has an electronegativity of 2.5, then the oxygen atom would have a slightly negative charge. The oxygen atom in  the carbon monoxide molecule would pull more electrons to its side since it has higher electronegativity making it slightly negative and the carbon would have a slightly positive charge as it would contain less electrons. This results to the formation of a polar molecule. A polar molecule is made when the molecule contains a slightly positive end and a slightly negative end. It would have a net dipole which is a result of the partial opposing charges in the molecule.
6 0
3 years ago
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