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yarga [219]
3 years ago
7

Please help... CaS cation and anion ZnCl2 cation and anion Cu2CrO4 cation and anion

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
7 0

Answer:Ionic compounds form when positive and negative ions share electrons and form an ionic bond. ... The positive ion, called a cation, is listed first in an ionic compound formula, followed by the negative ion, called an anion. A balanced formula has a neutral electrical charge or net charge of zero.

Explanation:Simple ions:  

Perchlorate ClO4- IO3-

Chlorate ClO3- BrO3-

Chlorite ClO2-  

Hypochlorite OCl- OBr-

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A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)
elixir [45]

Answer: \Delta H^{0}=-173.72 kJ/mol

Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}

Ca(OH)_{2}_{(aq)}+2HCl_{(aq)} → CaCl_{2}_{(s)}+2H_{2}O_{(l)}

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: \Delta H^{0}= -1002.82 kJ/mol

Calcium chloride: \Delta H^{0}= -795.8 kJ/mol

Water: \Delta H^{0}= -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]

-17.2=-1367.46+1002.82-2\Delta H

2\Delta H=17.2-364.64

\Delta H=-173.72

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

8 0
3 years ago
1. What do producers do, and what is another name for them?
Neko [114]

Producers are the foundation of every food web in every ecosystem—they occupy what is called the first tropic level of the food web. The second trophic level consists of primary consumers—the herbivores, or animals that eat plants. At the top level are secondary consumers—the carnivores and omnivores who eat the primary consumers. Ultimately, decomposers break down dead organisms, returning vital nutrients to the soil, and restarting the cycle. Another name for producers is autotrophs, which means “self-nourishers.” There are two kinds of autotrophs. The most common are photoautotrophs—producers that carry out photosynthesis. Trees, grasses, and shrubs are the most important terrestrial photoautotrophs.  In most aquatic ecosystems, including lakes and oceans, algae are the most important photoautotrophs.

8 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
4 years ago
What would the mass of 7 red apples be in kilograms?
Bogdan [553]

Answer:

I think it would be at 0.7 kg mass of the sun

8 0
3 years ago
A student following the reaction seen here calculated a theoretical yield of 38.3g C₆H₅Cl but when he did the experiment in the
Reil [10]

Answer:

96.1 %

Explanation:

Which teacher do you have lol

7 0
3 years ago
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