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2 years ago

1. Which of the following is a type of chemical used in fertilizer?

1 answer:

5 0

#1 is phosphate , #2 is pesticides, #3 is geometric settings, dominant water source as precipitation ,ground water , or surface water.and hydrobodynamics.#4 is microbiology.#5 is c, #6 is D,#7 is d

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

2 years ago

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one ele

liubo4ka [24]

Answer:

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms? It is a negative ion that has one more valence electron than a neutral bromine atom.

Explanation:

7 0

2 years ago

Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take

tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = 165°C

Depression in freezing point = $\Delta T_f=?$

$\Delta T_f=T- T_f=179^oC-165^oC=14^oC$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( organic compounds)

$\Delta T_f=14^oC$

$14^oC=1\times 40^oC kg/mol\times m$

$m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg$

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0

2 years ago

A molecule that has a central atom surrounded by three single bond pairs and one unshared pair would have a _____ shape

Fantom [35]

A molecule that has a central atom surrounded by three single bond pairs and one unshared pair would have a trigonal pyramidal shape. The electon arrangement of this is called tetrahedral. It involves one atom located at the apex and at the corners are three atoms with a trigonal base. An example would be ammonia or NH3. Nitrogen has five valence electrons so that it needs to three more electrons to satisfy the octet rule and be stable. It would share electrons with the three nitrogen present. In order, to achieve the most stable geometry, the three atoms of hydrogen would attach with a bond angle of 109 degrees.

6 0

2 years ago

Read 2 more answers

A student determines that a 9.8 g mixture of MgCl₂ (s) and NaNO3(s) contains 0.050 mol of

Damm [24]

mass MgCl₂ = mol x MM MgCl₂ = 0.05 x 95.211 g/mol = 4.76 g

mass Cl in MgCl₂ :

= (2 x AM Cl)/MM MgCl₂ x mass MgCl₂

= (2 x 35.5 g/mol)/95.211 g/mol x 4.76

= 3.55 g

% mass Cl in the mixture :

= (mass Cl / mass mixture) x 100%

= 3.55 / 9.8 x 100%

= 36.22%

6 0

1 year ago