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3 years ago

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the bonding of two amino acids to form a larger molecule requires the addition of nitrogen the release of carbon dioxide the rel

ease of a water molecule the addition of a water molecule

1 answer:

Ksivusya [100]3 years ago

3 0

What is the question here?

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What kingdom do fungi belong

miv72 [106K]

They’re Eukaryotic/ the Eukarya kingdom
Hope this helps :)

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2 years ago

Describe the motion of objects that are viewed from your reference frame both inside and outside while you travel inside a movin

Nutka1998 [239]

Answer:

2 c

Explanation:

8 0

3 years ago

You are provided with a compound fertilizer, 40-15-10.Calculate the quantity of fertilizer to add to a one hectare field supply

victus00 [196]

Answer:

a. 300 kg of Fertilizer

b. 225 kg of fertilizer

c.400 Kg of fertilizer

d.600 Kg of fertilizer

Explanation:

The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.

The percentages can be expressed as fractions as follows:

For nitrogen; 40/100 = 0.4

For phosphorus; 15/100 = 0.15

For potassium; 10/100 = 0.1

To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.

Quantity of fertilizer required to add to a hectare to supply;

a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer

b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer

c. Phosphorus at 60 kg/ha = 60/0.15 = 400 Kg of fertilizer

d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer

5 0

2 years ago

Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-

IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = $\frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M$

Concentration of $Br_{2}$ = $\frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M$

Concentration of NOBr = $\frac{0.102}{1}M=0.102M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106$

So, the reaction is not at equilibrium.

As $Q_{c}< K_{c}$ therefore reaction must run in forward direction to increase $Q_{c}$ and make it equal to $K_{c}$ .

4 0

2 years ago

reddi tstudent has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact me

prohojiy [21]

Explanation:

A point of temperature at which both solid and liquid state of a substance remains in equilibrium without any change in temperature then this temperature is known as melting point.

For example, melting point of water is $0 ^{o}C$ . So, at this temperature solid state of water and liquid state are present in equilibrium with each other.

Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is $215^{o}C$ then some of the solid will change into liquid state but the temperature will remains the same.

4 0

2 years ago