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2 years ago

A neighbor wants to use his backyard garden to

Chemistry

1 answer:

Anastaziya [24]2 years ago

4 0

Answer:

No, it is not

Explanation:

The reason for this answer is below

To conduct an experiment/investigation that involves the use of specific amounts of carbon dioxide and water by plants, one needs a controlled environment where the amount of carbon dioxide and water made available to the plant can be monitored and regulated as desired by the experimental design. The neighbor's backyard cannot provide this controlled environment and thus it is a bad idea.

Note that the neighbor's backyard cannot provide the controlled environment because the amount of carbon dioxide the plants will be exposed to there cannot be regulated as this is the carbon dioxide from the atmosphere and also the plants could get additional water from midnight dew which cannot be regulated also.

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Assuming an efficiency of 49.50%, calculate the actual yield of magnesium nitrate formed from 130.7 g of magnesium and excess co

Novay_Z [31]

Answer:

73.41 g

Explanation:

$Mg + Cu(NO_{3})_{3} ---> Mg(NO_{3})_{3} + Cu$

First find number of moles of Mg

n(Mg) $= \frac{mass}{molar mass} = \frac{130.7 g}{24.3050 g/mol}= 5.377 mol\\$

Second, find the number of moles of magnesium nitrate produced by this number of moles of magnesium.

The molar ratio is one is to one (using stoichiometric coefficients from the balanced equation). So 5.377 mol of magnesium produces 5.377 mol of magnesium nitrate

Third, find what mass of magnesium nitrate that is

mass(Mg(NO3)2) = mol * molar mass

= 5.377 mol * 148.3148 g/mol

= 148.3148 g

But the system is inefficient. It only has an efficiency of 49.50 so what will be the actual, not theoretical yield.

actual yield = efficiency * theoretical yield

= 49.5% * 148.3148 g

= 73.41 g

8 0

2 years ago

In the following equation, which element has been oxidized?

Bumek [7]

The element that has been oxidized is Al ( Answer A)

Explanation

if the oxidation number of an element is greater in the product side than the reactant side then the element has lost the electrons . This mean that the element has been oxidized.

Al is oxidized because it move from oxidation state Zero(0) in reactant side to oxidation state 3⁺ in product side.

This implies that Al loses 3 electrons.

6 0

2 years ago

Is the following change an oxidation or reduction?

AnnyKZ [126]

Answer:

12113

Explanation:

step by step:no

8 0

2 years ago

The highly reactive elements in group 7A are known for forming salts. What are they called? A. Transition metals B. Metalloids C

madam [21]

The answer is C Halogens.

8 0

2 years ago

In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,

Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$K_{sp}=s\times {2s}^2$

$K_{sp}=4s^3$

Given s = 7.4×10⁻³ M

So, Ksp is:

$K_{sp}=4\times (7.4\times 10^{-3})^3$

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

s' = 4.0×10⁻⁵ M

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

3 0

2 years ago