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Leona [35]
3 years ago
9

A. 2 pts: In terms of conservation of mass, what were the results of each experiment (which one conserved mass and which one did

not)?
B. 4 pts: PROVIDE a SPECIFIC explanation for "why" the results in both videos were different.
. 2 pts: Discuss differences in the experimental steps between the two videos (for example what was done in video one that was not done in
video two, why would this make a difference in conserving mass?)
• 2 pts: Look at both chemical equations (below for convenience), what difference in the type of products would have contributed to the
differences in why one conserved mass and one did not?
REACTION 1: Pb(NO3)2 (aq) + 2K1 (aq) --> 2KNO3 (aq) +Pölz (s)
• REACTION 2: NaHCO3(s) + CH3COOH(aq) → CO2(g) + H2Om + CH3COONa(aq)

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Answer:

dftdffdgxrdfdrdxxxxdddddddddddddeeertdffddy

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A compound has 15.39 g of gold for every 2.77 g of chlorine.
Ugo [173]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The question is to determine the simplest mass ratio of gold to Chlorine in the compound.

Since the mass of gold in the compound compared to chlorine is 15.39 g for every 2.77 g, the mass of gold per gram of chlorine is given as:

15.39 / 2.77 = 5.56 g of gold to two decimal places

<em>Therefore, for every 5.56 g of gold, there is 1 g of chlorine.</em>

<em>Note : The ratio in which different elements combine by mass to form a compound is given by the law of constant composition which states that, "all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass.</em>

3 0
3 years ago
3 electron layers in the electron shell and 1 electron in the last electron layer​
Nikolay [14]

Answer:

The closest shell to the nucleus is called the "1 shell" (also called the "K shell"), followed by the "2 shell" (or "L shell"), then the "3 shell" (or "M shell"), and so on farther and farther from the nucleus.

8 0
3 years ago
Which of the following is kept constant when using a bomb calorimeter?
trapecia [35]
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.

6 0
4 years ago
If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need
dimulka [17.4K]
In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps
5 0
3 years ago
when carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4g of carbon were burned in the presence of
SpyIntel [72]
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
7 0
3 years ago
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