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rusak2 [61]
3 years ago
6

____ is the transfer of energy in the form of rays or waves. a. Conduction c. Radiation b. Convection d. Condensation  

Chemistry
2 answers:
Maksim231197 [3]3 years ago
5 0

Explanation:

When energy is emitted or transmitted in the form of waves or rays that travel through the space or material medium then this energy is known as radiation.

For example, Sum emits radiation in the form of heat, light and particles.

Thus, we can conclude that out of the given options, radiation is the transfer of energy in the form of rays or waves.

NemiM [27]3 years ago
4 0
Radiation <span>is the transfer of energy in the form of rays or waves. </span>
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Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
2 years ago
The chemical equation shows the production of magnesium chloride.
Norma-Jean [14]

Answer:

1 gram of H2 will be produced from 12 grams of Mg.

Explanation:

According to Stoichiometry, 0.5 moles of Mg are present. 1 mole of Mg produces 1 mole of H2, so 0.5 moles of Mg will produce 0.5 moles of H2. Multiplying molar mass of H2 i.e. 2 gram/mole with 0.5 moles, we can find the mass of H2 in grams which is 1 gram.

3 0
3 years ago
An alloy of bronze is manufactured by melting 51.2 g of copper with 6.84 g of tin. What is the percent copper in the bronze?
Harlamova29_29 [7]

Answer:

Percentage of copper = 88%

Explanation:

Given data:

Mass of copper = 51.2 g

Mass of tin = 6.84 g

Percentage of copper = ?

Solution:

Formula:

Percentage of copper = mass of copper / total mass × 100

Now we will determine the total mass:

Total mass = mass of copper + mass of  tin

Total mass = 51.2 g + 6.84 g

Total mass = 58.04 g

Now we will calculate the percentage of copper.

Percentage of copper = 51.2 g / 58.04 g × 100

Percentage of copper = 0.88 × 100

Percentage of copper = 88%

5 0
3 years ago
Four molecules have carbon as the central atom and hydrogen atoms around it. Which shape would give rise to a polar molecule?
zzz [600]
Four molecules have carbon as the central atom and hydrogen atoms around it. The shape that would give rise to a polar molecule is a bent shape. <span>The bent shape will not be symmetrical which will result to being polar. Hope this answers the question.</span>
7 0
2 years ago
Show your work with good use of units, rounding, and significant figures. [Hint: it is good practice to show the value of your a
mel-nik [20]

Heat required : 4.8 kJ

<h3>Further explanation </h3>

The heat to change the phase can be formulated :

Q = mLf (melting/freezing)

Q = mLv (vaporization/condensation)

Lf=latent heat of fusion

Lv=latent heat of vaporization

The heat needed to raise the temperature

Q = m . c . Δt

1. heat to raise temperature from -20 °C to 0 °C

\tt Q=10\times 2.09\times (0-(-20)=418~J

2. phase change(ice to water)

\tt Q=10\times 333=3330~J

3. heat to raise temperature from 0 °C to 25 °C

\tt Q=10\times 4.18\times (25-0)=1045~J

\tt Q~tot=418+3330+1045=4793~J\rightarrow rounding~and~2~sig~figs=4.8~kJ

3 0
2 years ago
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