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My name is Ann [436]
2 years ago
8

9) A solid magnesium flare has a mass of 2300 g and a volume of 703 cm3. What is the density of the

Chemistry
1 answer:
harkovskaia [24]2 years ago
8 0

Answer:

siks creo que creo que siks

You might be interested in
In the chemical reaction:
velikii [3]

Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Cu + 2 AgNO₃  → 2 Ag + Cu(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Cu: 1 mole
  • AgNO₃: 2 moles
  • Ag: 2 moles
  • Cu(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?

amount of moles of AgNO_{3} =\frac{1.8 moles of Cux2 moles of AgNO_{3} }{1 mole of Cu}

<u><em>amount of moles of AgNO₃= 3.6 moles</em></u>

But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.

<h3>Summary</h3>

In summary, AgNO₃ will be the limiting reagent.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

8 0
1 year ago
A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0
3 years ago
In a heating curve, when is the temperature constant? need answers!
Varvara68 [4.7K]
During a phase change
3 0
3 years ago
Read 2 more answers
II. Ionic Equations
mario62 [17]

Answer:

Complete ionic: \begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Net ionic: \begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

Explanation:

Start by identifying species that exist as ions. In general, such species include:

  • Soluble salts.
  • Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2. These three salts will exist as ions:

  • Each \rm AgNO_3\, (aq) formula unit will exist as one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion.
  • Each \rm CaCl_2 formula unit will exist as one \rm Ca^{2+} ion and two \rm Cl^{-} ions (note the subscript in the formula \rm CaCl_2\!.)
  • Each \rm Ca(NO_3)_2 formula unit will exist as one \rm Ca^{2+} and two \rm {NO_3}^{-} ions.

On the other hand, \rm AgCl is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2 (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each \rm AgNO_3\, (aq) formula unit will exist as only one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion. However, because the coefficient of \rm AgNO_3\, (aq)\! in the original equation is two, \!\rm AgNO_3\, (aq) alone should correspond to two \rm Ag^{+}\! ions and two \rm {NO_3}^{-}\! ions.

Do not rewrite the salt \rm AgCl because it is insoluble.

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of \rm Ca^{2+} and two units of \rm {NO_3}^{-}. Doing so will give:

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}.

Simplify the coefficients:

\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

7 0
2 years ago
When nitrogen gas reacts with hydrogen gas, ammonia gas is formed. how many grams of hydrogen gas are required to react complete
aniked [119]
The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given 
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol 
then 3.5 L of gas is occupied by - 3.5 L / 22.4 L/mol = 0.16 mol 
number of moles of N₂ present - 0.16 mol
1 mol of N₂ requires 3 mol of H₂
therefore 0.16 mol of N₂ requires - 3 x 0.16 = 0.48 mol of H₂
mass of H₂ required - 0.48 mol x 2 g/mol = 0.96 g
0.96 g of H₂ is required
3 0
3 years ago
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