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3 years ago

Chemicals that control weeds are called

Chemistry

2 answers:

mote1985 [20]3 years ago

6 0

Answer:

B. herbicides

Explanation:

it kills weeds.

DIA [1.3K]3 years ago

5 0

Answer:

The answer is D

Explanation:

because insecticides are used for insects but pesticides are used to kill off unwanted plants

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Fungi is an example of a decomposer.

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What is true of enzymes? They only speed up reactions. They speed up or slow down any reaction in a living thing. They act as ca

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They act as catalysts

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What procedure could a student use to examine an intensive property of a rectangular block of wood

photoshop1234 [79]

An intensive property does not change when you take away some of the sample. The procedures that a student could use to examine the intensive property of a rectangular block of wood are the hardness, color, density and molecular weight.

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4 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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3 years ago

A compound that is usually used as a fertilizer can also be used as a powerful explosive. the compound has the composition 35.00

maxonik [38]

Answer:

Empirical Formula = NH₄NO₃ (Ammonium Nitrate)

Solution:

Step 1: Calculate Moles of each Element;

Moles of N = %N ÷ At.Mass of N

Moles of N = 35.0 ÷ 14

Moles of N = 2.5 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 59.96 ÷ 16

Moles of O = 3.7475 mol

Moles of H = [100% - (%N + %O)] ÷ At.Mass of H

Moles of H = [100% - (35.0 + 59.96)] ÷ 1.008

Moles of H = [100% - 94.96] ÷ 1.008

Moles of H = 5.04 ÷ 1.008

Moles of H = 5 mol

Step 2: Find out mole ratio and simplify it;

N H O

2.5 5 3.7475

2.5/2.5 5/2.5 3.7475/2.5

1 2 1.5

Multiply Mole Ratio by 2,

2 4 3

Result:

Empirical Formula = N₂H₄O₃

Or,

Empirical Formula = NH₄NO₃

This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.

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3 years ago