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3 years ago

In order to predict whether or not a single replacement reaction takes place you need to consult a chart that shows the

1 answer:

6 0

Activity series of metals. it is a series of metals based on their reactivity from highest to lowest. Potassium occupies the top most level of reactivity. The reactivity means the ability to displace hydrogen gas from water and acid solutions.

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Apart from plants, what other group of organisms photosynthesises?

Licemer1 [7]

Answer:

Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis

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2 years ago

Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is . The organi

nevsk [136]

Explanation:

Haemoglobin consists of heme unit which is comprised of an $Fe^{2+}$ and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.

The prosthetic group of hemoglobin and myoglobin is - Heme

The organic ring component of heme is - Porphyrin

Under normal conditions, the central atom of heme is - $Fe^{2+}$

In deoxyhemoglobin , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.

The central atom has six bonds: four to nitrogen atoms in the porphyrin, one to a histidine residue, and one to oxygen.

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2 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

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3 years ago

How much heat energy is required to raise the temperature of 0.358 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co

NNADVOKAT [17]

In order to calculate how much heat is needed to raise the temperature you need to use the formula q =mass x specific heat x (final temperature- initial temperature) where q represents heat being absorbed or released. Before you begin you would convert kg to g because the specific heat is measure in g. So you would set up the equation as q = 358 g x .092 x (60-23 degrees Celsius) which would give you 1218.6

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3 years ago

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Keith_Richards [23]

Answegood

Explanation:this answer is two it has letters

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2 years ago

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