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LekaFEV
1 month ago
12

Which of the following statements is true about one formula unit of RuF2?​

Chemistry
1 answer:
KiRa [2.9K]1 month ago
6 0
Para mí, la respuesta es E. Está formada por un ion Ru2+ y dos iones F-.
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The density of a 50% solution of naoh is 1.525 g/ml. what volume of a solution that is 50% by weight naoh is required to make 0.
eduard [2782]
We assume that the stated 50% is measured by volume. Molarity defines the concentration in terms of moles of solute per volume of solution.

      To find the moles of NaOH, use: (0.1 moles / L)(0.4 L)
                      n = 0.04 moles of NaOH

Assuming we start with 1 mL of 50% NaOH solution, 
  
                        (1 mL solution)(1.525 g/mL)(0.50) = 0.7625 g
Then, the number of moles calculates as follows,[
                  0.7625 g NaOH x (1 mol / 40 g) = 0.01906 moles of NaOH

                The volume of solution required can be determined by:(0.04 moles of NaOH)(1 mL solution / 0.01906 moles of NaOH)
                 
                 Thus, the needed volume comes out to be 2.09 mL

Answer: 2.09 mL
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29 days ago
Persamaan setara pada reaksi besi dengan asam klorida membentuk besi (II) klorida dan gas hidrogen
VMariaS [2998]
<span>Reaksi antara besi dan asam klorida menghasilkan besi (II) klorida serta gas hidrogen.</span>
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1 month ago
When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro
VMariaS [2998]
To tackle this problem, one must first determine the specific heat of water, which is the energy required to raise the temperature of 1 g of water by 1 degree C. The relationship is given by the formula q = c X m X delta T, where q indicates the specific heat of water, m signifies the mass, and delta T denotes the temperature change. The specific heat of water is 4.184 J/(g X degree C). The temperature of the water increased by 20 degrees, therefore: 4.184 x 713 x 20.0 = 59700 J, rounded to 3 significant digits, equals 59.7 kJ. This value indicates the energy required to produce B2O3 from 1 gram of boron. To convert this to kJ/mole, additional calculations are required. The gram atomic mass of Boron is 10.811, so dividing 1 gram of boron by 10.811 results in.0925 moles of boron. Given that 2 moles of boron are needed for the formation of 1 mole of B2O3, dividing the moles of boron by two yields.0925/2 =.0462 moles. Consequently, dividing the energy in KJ by the number of moles provides KJ/mole: 59.7/.0462 = 1290 KJ/mole.
7 0
29 days ago
Upon heating with acid salicylic acid can form a polymer. what is its structure likely to be
KiRa [2933]
Refer to the attached image. This does not depict the polymerization reaction of salicylic acid. It merely illustrates the structure of salicylic acid and what it looks like post-polymerization. The polymerization occurs through the linking of salicylic acid molecules, resulting in the formation of one water molecule.

4 0
23 days ago
2CH4(g)⟶C2H4(g)+2H2(g)
Alekssandra [3086]

Answer: The enthalpy change for the reaction is, 201.9 kJ

Explanation:

Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.

This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.

The balanced equation for CH_4 appears as follows,

2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)    \Delta H^o=?

The intermediate balanced reactions are outlined as follows,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890.3kJ

(2) C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)     \Delta H_2=-136.3kJ

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=-571.6kJ

(4) 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)     \Delta H_4=-3120.8kJ

Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:

(1) 2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)     \Delta H_1=2\times (-890.3kJ)=-1780.6kJ

(2) C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)    \Delta H_2=-(-136.3kJ)=136.3kJ

(3) H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ

(4) 2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)     \Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ

Therefore, the expression for the enthalpy of the reaction is,

\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)

\Delta H=201.9kJ

Hence, the enthalpy change for this reaction is, 201.9 kJ

5 0
1 month ago
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