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kari74 [83]
3 years ago
5

Helppppppppppppppppppppppppp

Chemistry
1 answer:
sergey [27]3 years ago
8 0

Answer:

lose 1 2+

Explanation:

I honestly don't know

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Solid cadmium sulfide reacts with an aqueous solution of sulfuric acid . Express your answer as a balanced chemical equation. Id
max2010maxim [7]

Explanation:

When solid cadmium sulfide reacts with an aqueous solution of sulfuric acid then the reaction will be as follows.

          CdS(s) + H_{2}SO_{4}(aq) \rightarrow CdSO_{4}(aq) + H_{2}S(g)

Hence, ionic equation for this reaction is as follows.

      CdS(s) + 2H^{+}(aq) + SO^{2-}_{4}(aq) \rightarrow Cd^{2+}(aq) + SO^{2-}_{4}(aq) + H_{2}S(g)

Therefore, net ionic equation for this reaction is as follows.

      CdS(s) + 2H^{+}(aq) \rightarrow Cd^{2+}(aq) + H_{2}S(g)

8 0
3 years ago
Which is a drawback of agricultural technology?
Umnica [9.8K]

Answer:

B.

Explanation:

Everything else is not a drawback, it is a benefit.

4 0
3 years ago
6th grade science help me if you can c:
lesya [120]
B. Troposphere, stratosphere, mesosphere, thermosphere
4 0
3 years ago
Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is described by the following thermoche
vredina [299]

Answer:

-537.25 kJ/mol is the standard enthalpy of formation of solid glycine.

Explanation:

4C_2H_5O_2N(s) + 9O_2(g)\rightarrow 8CO_2(g) + 10H_2O(l) + 2N_2(g) ,\Delta H^{rxn} =-3896 kJ/mol

Standard enthalpy of formation of oxygen gas= \Delta H_{f,O_2}=0

Standard enthalpy of formation of carbon dioxide= \Delta H_{f,CO_2}=-393.5 kJ/mol

Standard enthalpy of formation of water = \Delta H_{f,H_2O}=-285.8 kJ/mol

Standard enthalpy of formation of nitrogen gas= \Delta H_{f,N_2}=0

Standard enthalpy of formation of glycine = \Delta H_{f,gly}=?

Enthalpy of the reaction = H_{rxn}=-3896 kJ/mol

H_{rxn} =

=8\times \Delta H_{f,CO_2} +10\times \Delta H_{f,H_2O}+2\times \Delta H_{f,N_2} - (4\times \Delta H_{f,gly}+9\times \Delta H_{f,O_2})

-3896 kJ/mol=8\times (-393.5 kJ/mol)+ 10\times (-285.8 kJ/mol)+0 - (4\times \Delta H_{f,gly} +0)

On rearranging :

4\times \Delta H_{f,gly}=8\times (-393.5 kJ/mol)+ 10\times (-285.8 kJ/mol)+ 3896 kJ/mol

\Delta H_{f,gly}=\frac{-2149 kJ/mol}{4}=-537.25 kJ/mol

-537.25 kJ/mol is the standard enthalpy of formation of solid glycine.

6 0
3 years ago
Determine the molarity for a solution of na3po4 prepared by diluted 10.0 ml of a 0.33 m solution of na3po4 to 1.00 l. 2. if 5.00
AVprozaik [17]
Answer i) The molarity for the solution of Na_{3} PO_{4} prepared after dilution from 10 mL of 0.33 M solution to 1 L solution is 3.3 X 10^{-3},

m _{1}V_{1}=m_{2} V_{2}

 by using this formula. 

So, Molarity of unknown solution = (10 X 0.33) / 1000 = 3.3 X 10^{-3}.

Answer ii) When you dilute the solution prepared in question (1 ) to 50.0 mL from 5.00 mL of 1L solution then, you should take an aliquot of 5 mL and then dilute upto 50 mL that will be the result of another 10 times dilution which will result in the new concentration as 0.00033 M for Na_{3} PO_{4} solution.

Answer iii)
Question iii is incomplete. Complete question is attached below;
Solution:
Given: A = 0.982, b = 1 cm, c = 0.0001 M

we know that, A = ebc
Therefore, ∈ = A/bc = 0.982/(1 X 0.0001) = 9820 M-1 cm-1.

Thus, molar absorptivity of solution is 9820 M-1 cm-1.

8 0
3 years ago
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