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dimulka [17.4K]
3 years ago
13

What type of chemical reaction might be confused with the synthesis reaction in part 1, and how can you distinguish these two re

actions? (b) write a balanced chemical equation for the synthesis reaction in part 1?
Chemistry
1 answer:
Sergio039 [100]3 years ago
4 0
I as m not sure but researching it and see what you find
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Which one is it ? I need help
Lostsunrise [7]
It would 3 because if you read it tells you the answer
8 0
3 years ago
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In terms of energy, how would you classify the following chemical reaction? 2CuO + 315 kJ → 2Cu + O2
Snowcat [4.5K]
"Endothermic" is the way among the following choices given that you would <span>classify the following chemical reaction, in terms of energy. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has come to your help.</span>
6 0
3 years ago
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A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.
Sergio039 [100]

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

                F_{net} = Buoyancy force(F_{b}) - weight force(w)

Also, buoyancy force (F_{b}) = (volume submerged in water × density of water) + (volume in oil × density of oil)

          (F_{b}) = (0.592 V \times \rho) + (1 - 0.592)V \times 1000 g          

                      = (0.592 V \times \rho + 408 V) g

As,   W = V × density of graphite × g

It is given that density of graphite is 2.16 g/cm^{3} or 2160 kg/m^{3}.

So, W = 2160 V g

F_{net} = (0.592 V \rho + 408 V) g - 2160 V g = 0

            0.592 \rho = 1752

     \rho = 2959.46 kg/m^{3} or 2.959 g/cm^{3} is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 cm^{3} oil = 35.3 \times 2.959 104.7 g

Hence, in second weighing total mass will be calculated as follows.

                       (124.8 + 104.7) g

                       = 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

5 0
3 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
3 years ago
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