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3 years ago

Which metals have similar chemical properties? Check all that apply. barium

Chemistry

1 answer:

rodikova [14]3 years ago

6 0

Alkina metals all have the similar proteries

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What suffix do the halogen group use

sesenic [268]

Answer: That is, the suffix of the compound is unchanged by the presence of the halogen, and the halogen is included as a prefix in the name. In acyl halides, the suffix -oyl chloride is appended to the name. For example, CH3CH2COCl is called propanoyl chloride.

Explanation:

6 0

2 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.

Helen [10]

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

Q: heat absorbed

C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)

ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

3 0

3 years ago

I need to know the following above

arsen [322]

3Zn + 8HNO3 ---> 3Zn(NO3)2 + 4H2O + 2NO IF IT IS COLD AND DILUT NITRIC ACID .

IF IT IS HOT AND CONCENTRATED THEN:

Zn+ 4HNO3 ---> Zn(NO3)2 +2H2O +2NO2

6 0

3 years ago

How many variables are there in science

Sedaia [141]

There are three variables independent, dependent ,and controlled

7 0

3 years ago