What is the total heat energy required to

Dose All plants have a vascular system of pipes that transport fluids]

USPshnik [31]

Answer:

certain types of plants (vascular plants) have a system for transporting water, minerals

4 0

3 years ago

How can we increase the rate of collisions between the reactants in this reaction? mg 2hcl → mgcl2 h2

Minchanka [31]

The frequency of collisions between the two reactants increases as the concentration of the reactants increases. When collisions happen, they don't always cause a reaction (atoms misaligned or insufficient energy, etc.). Higher concentrations result in more collisions and reaction opportunities.

Increasing a reactant's surface area increases the frequency of collisions and thus the reaction rate. The surface area of several smaller particles is greater than that of a single large particle. The greater the available surface area for particles to collide, the faster the reaction will occur.

<h3>How does concentration affect the rate of collisions between reactants?</h3>

Thus, we can conclude that by increasing the concentration of Mg in the reaction mixture we increase the rate of collisions between the reactants in this reaction.

<h3>What does the half reaction of an oxidation-reduction reaction show?</h3>

Iron gains electrons in the half reaction of an oxidation-reduction reaction. What does iron's electron gain mean? It has been reduced. Predict the product that will precipitate out of the reaction using the solubility rules and the periodic table.

Learn more about collisions of particles:

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4 0

2 years ago

Why is it important for an engineer who is designing a new home-cooling/heating system to have knowledge about the transfer of h

Masteriza [31]

Cooling and heating system involves the collection and dissipation of heat appropriately. Heat gradient is important for heat transfer between systems because heat travels from the warmer system to the cooler system. The right medium for heat transfer is also important for this exchange – A good conductor of heat is required.

4 0

3 years ago

What group is the element barium(Ba) In? A. 1 B. 6 C. 56 D. 4

LenaWriter [7]

Ummmmmmmmmmm 56 yes 56

7 0

3 years ago

What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of

weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{31.862\ g}{63.546\ g/mol}$

Moles of copper = 0.5014 moles

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 284\ g$ = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{193.12\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.0649 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.0028 moles

Given:

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.35 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K

⇒V = 25.76 L

4 0

3 years ago