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julia-pushkina [17]
3 years ago
9

How do you write a chemical equation given a word equation ? (Chemistry)

Chemistry
1 answer:
earnstyle [38]3 years ago
5 0
Step one: Identify reactants and products and place them in a word equation.

Step two: Convert the chemical names into chemical formulas. Place them based on the chemical equation and write the state symbols.

Step three: balance the chemical equation.
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What controls flowering in many<br> plants?
katrin2010 [14]

Answer:

All flowering is regulated by the integration of environmental cues into an internal sequence of processes. These processes regulate the ability of plant organs to produce and respond to an array of signals. The numerous regulatory switches permit precise control over the time of flowering.

Explanation:

3 0
3 years ago
rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0
3 years ago
Read 2 more answers
9A. A sample of hydrogen at 1.56 atm had it's pressure decreased to 0.73
Jlenok [28]

Answer:idk

Explanation:

Idk

8 0
3 years ago
How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*
MrMuchimi

<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L

Hence, 6.57 L of solution can be made.

5 0
3 years ago
A student uses 5g of carbon to extract copper metal from its ore:
Umnica [9.8K]

Answer:

drt4gxhknwo

Explanation:

4 0
2 years ago
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