1. carbon dioxide is a gas at low temperatures at mars because in the oxidizing environment organic compounds are oxidized to form carbon dioxide.
2. The 1 electron in outermost shell of Na is shared with 7 electrons of outermost shell of chlorine giving neutral charge on compound.
3. Electrostatic force of attraction is between the metal and non-metal.
4. When metal and non metal exchange electrons to form a neutral or no charge compound it is said to have form ionic bonds.
Explanation:
1. Temperature at Mars is very low -80 Fahrenheit (-60 degrees) because water is not present in the planet. Carbon dioxide is abundant in Mars. The atmosphere is oxidizing at Mars which oxidizes the organic compounds and forms carbon dioxide.
2. NaCl combines by sharing of electrons forming ionic bonding. Different atoms of the different element share electrons to form ionic bonds. Such bond is formed when electrons is transferred between the atoms. In the NaCl, Na has 1 electron (electropositive) in its outer shell and chlorine has 7 electrons (electronegative). Both share the electrons getting their octet complete and a neutral charge on the compound formed.
3. Electrostatic force of attraction is between the metal and non-metal when bond is formed. The ionic bonds is formed between metal and non metals when electron exchange takes place. The electrostatic force is the attraction between two opposite charges on the ion.
4. When a metal and non metal exchange electrons in which metal is electropositive and non metal is electronegative the bonds form is called ionic bond. The electron is transferred from metal to non metal and thus giving neutral charge on the compound i.e. the outer shell has its octet complete.
Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. Although these two aspects of bomb calorimetry make for accurate results, they also contribute to the difficulty of bomb calorimetry. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.
Introduction
Calorimetry is used to measure quantities of heat, and can be used to determine the heat of a reaction through experiments. Usually a coffee-cup calorimeter is used since it is simpler than a bomb calorimeter, but to measure the heat evolved in a combustion reaction, constant volume or bomb calorimetry is ideal. A constant volume calorimeter is also more accurate than a coffee-cup calorimeter, but it is more difficult to use since it requires a well-built reaction container that is able to withstand large amounts of pressure changes that happen in many chemical reactions.
Most serious calorimetry carried out in research laboratories involves the determination of heats of combustion ΔHcombustion" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.4px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ΔHcombustionΔHcombustion, since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, and this accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following:
Steel bomb which contains the reactantsWater bath in which the bomb is submergedThermometerA motorized stirrerWire for ignition
is usually called a “bomb”, and the technique is known as bomb calorimetry
Another consequence of the constant-volume condition is that the heat released corresponds to qv , and thus to the internal energy change ΔUrather than to ΔH. The enthalpy change is calculated according to the formula
(1.1)ΔH=qv+ΔngRT" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.4px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; width: 10000em !important; position: relative;">ΔH=qv+ΔngRT(1.1)(1.1)ΔH=qv+ΔngRT
Δng" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 14.4px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ΔngΔng is the change in the number of moles of gases in the reaction.
<span>density is how much matter is packed into one area. so the higher density the more likely it is to be heavier. for example, cork has a very low density and iron has a higher density.</span>
Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
Hi Sydney!
I can't draw in this question, but there is a picture showing this phase for you to follow when you draw it.
Hope This Helps :)
