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Vladimir [108]
3 years ago
12

The table shows data for two groups of plants one grown with fertilizer and the other without fetalizer was the mean height of t

he plant given fertilizer​
Chemistry
2 answers:
Yuki888 [10]3 years ago
7 0
Do you need variables orrr what is the question ?
AleksAgata [21]3 years ago
6 0

Answer:

What is the question?

Explanation:

You might be interested in
1. What is the best means to separate ammonium chloride, sand, and zinc chloride? And a
ozzi

1. A mixture of ammonium chloride, sand, and zinc chloride should be separated by sublimation.

2. A mixture of zinc chloride and silver chloride should be separated through crystallization.

<h3>What is a separation technique?</h3>

A separation technique can be defined as a technique that is typically used to separate or convert two (2) or more mixture and solution of chemical substances into distinct product such as chemical compounds or elements.

<h3>The types of separation technique.</h3>

In Chemistry, there are various types of separation technique used for the separation of mixtures or solutions and these include:

  • Crystallization
  • Fractional distillation
  • Filtration
  • Evaporation
  • Sublimation
  • Decantation
  • Simple distillation
  • Chromatography

In this scenario, the most effective and efficient means to separate a mixture of ammonium chloride, sand, and zinc chloride is by sublimation from solid to gas state.

On the other hand, the best means to separate a mixture of zinc chloride and silver chloride is through crystallization.

Read more on crystallization here: brainly.com/question/4980962

7 0
2 years ago
2 Points
pishuonlain [190]

Answer:

B

Explanation:

Boyle was known for rejecting Aristotle's theory basing it on the four elements (earth, air, fire, water).

4 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
3 years ago
A 1.250-g sample of benzoic acid, C7H6O2, was placed in a combustion bomb. The bomb was filled with an excess of oxygen at high
Degger [83]

Answer:

3224 kJ/mol

Explanation:

The combustion of benzoic acid occurs as follows:

C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE

The change in temperature in the reaction is the change due the energy released, that is:

3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts

To find the heat released per mole we have to find the moles of benzoic acid:

<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>

1.250g * (1mol / 122.12g) = 0.0102 moles

<em />

The dE combustion per mole of benzoic acid is:

33.00kJ / 0.0102moles =

<em>3224 kJ/mol </em>

4 0
3 years ago
3. A student decides to determine the elemental identity of a metal
miss Akunina [59]

Answer:

part (a) was used to calculate the molar mass of the metal M. In part (c) students proposed an experimental test that could be used to determine if MI 2 is an ionic solid, and then explained how the results of the test supported this claim. Students explained why I 2 is solid at room temperature, but Br 2 is a liquid in part (d).  its {c}

5 0
3 years ago
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