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nasty-shy [4]
3 years ago
14

Balance this chemical equation __K+__H2O>>> __KOH+__H2

Chemistry
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

2K+2H2O=2KOH+H2

Explanation:

Try to ensure there is same number of reactants as products

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The equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of ox
Vaselesa [24]
<span>0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles</span>
7 0
3 years ago
2a. Density is the relationship between mass and volume. Density has
Charra [1.4K]

The volume of copper : 3.24 ml

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}

ρ = density , g/cm³ or kg/m³

m = mass , g or kg

v = volume , cm³  or m³

A common example is the water density of 1 gr / cm³

The density of copper : 8.96 gr/ml

mass of copper : 29 g

then the volume :

\tt V=\dfrac{m}{\rho}\\\\V=\dfrac{29~g}{8.96~g/ml}\\\\V=\boxed{\bold{3.24~ml~or~3.24~cm^3}}

8 0
3 years ago
Hemoglobin (Hb) can be viewed as having two quaternary states, a low oxygen affinity state (T), and a high oxygen affinity state
natta225 [31]
<h3><u>Answer;</u></h3>

The statements that are True are;

  • Upon binding a molecule of oxygen, Hb undergoes a conformational change that makes the binding of subsequent O2 molecules easier.
  • The conformational change induced in Hb upon binding oxygen is the result of a small movement (0.2 Å) of the iron cation in the center of heme.
  • Site-directed mutagenesis studies have indicated that the cooperativity of O2 binding in Hb is attributable to the movement of the F helix in Hb.
<h3><u>Explanation</u>;</h3>
  • Hemoglobin is a key pigment in the blood that transports oxygen gas to all the tissues in the body. It is made up of two types of chains; that is two alpha chains and two beta chains.  
  • in its deoxygenated state hemoglobin has a low affinity for oxygen compared to myoglobin. When oxygen is bound to the first subunit of hemoglobin it leads to subtle changes to the quaternary structure of the protein. This in turn makes it easier for a subsequent molecule of oxygen to bind to the next subunit.
8 0
3 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
How many moles of KClO3 are produced from 25 moles of KCL
e-lub [12.9K]

Answer: 25

Explanation:

8 0
3 years ago
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