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vagabundo [1.1K]
4 years ago
15

Chemicals used for controlling insects and weeds could reduce soil fertility by:

Chemistry
2 answers:
Vikki [24]4 years ago
8 0
All chemicals have harmful properties which also effect the stuff around it like weed killer even though it kills weeds the chemical still stays in the ground which makes it harmful for other plants
dimaraw [331]4 years ago
6 0

When farmers use chemicals to kill some weed or bugs, those chemicals also were absorbed by the plant and the soil.

That absorption can affect fertility because the chemical can change some chemical and physical traits from the soil and plant.

In worse cases, it can cause erosion, this means,  that the soil is "dead" and nothing will grow.

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What does a argon bohr model look like
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This is bohr model of argon




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3 years ago
A sample of an unknown metal has a mass of 6.557 g. The metal was carefully added to a graduated cylinder containing 10.50 mL of
Levart [38]

Answer:

Explanation:

Be careful. The tricky part of the problem is that there are 4 places of sig digs.

m = 6.557 grams

V = 11.16 - 10.50 = 0.66

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density = 6.557/0.66 = 9.935 g/mL

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4 years ago
Consider the procedures in a simple experiment used to demonstrate the behavior of gases
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B, C, E

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Which property of matter is conserved in chemical reactions and shown by balanced equations? mass volume density shape
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3 0
4 years ago
Read 2 more answers
A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
irinina [24]

Answer:

m_{KBr}=6.030gKBr

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

(T_b-T_b_0)=i*m*K_b

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\  \\m=0.507mol/kg

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

n_{KBr}=0.507mol/kg*0.1kg=0.0507mol

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr

Best regards.

3 0
4 years ago
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