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3 years ago

Balance the following equations

Chemistry

2 answers:

Svetlanka [38]3 years ago

4 0

ANSWER

<em>2Ba + 2HBr → 2BaBr + H2</em>

<em>2BiCl3 + 3H2S → Bi2S3 + 6HCl</em>

<em>Br2 + 2KI → I2 + 2KBr</em>

<em>4Fe + 3O2 → 2Fe2O3</em>

<h3>I HOPE THIS WILL HELP YOU IF NOT THEN SORRY HAVE A GREAT DAY:)</h3>

allochka39001 [22]3 years ago

4 0

Answer:

$3Mg _{(s)} + N _{2(g)} → Mg_{3} N _{2(s)}$

$Ba _{(s)} + 2HBr_{(g)} → BaBr _{2(s)} + H _{2(g)}$

$2BiCl_{3(s)} + 3H_{2} S _{(g)} → Bi _{2} S _{ 3(g)} + 6HCl _{(g)} \\$

$Br_{2(g)} + 2KI _{(g)} → I _{2(g)} +2 KBr _{(g)}$

$4Fe_{(s)}+ 3O_{2(g)} → 2Fe_{2} O _{3(s)}$

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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.

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4 years ago

A more reactive metal displaces a less reactive metal from salt solution. Such reactions are called:

motikmotik

Answer:

Displacement Reaction.

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3 years ago

Read 2 more answers

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl

nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of caffeine extracted is $P = 39.8 \ mg$

Explanation:

From the question above we are told that

The K value is $K = 4.6$

The mass of the caffeine is $m = 40 mg$

The volume of water is $V = 2 mL$

The volume of caffeine is $v_c = 2 mL$

The number of times the extraction was done is n = 3

Generally the mass of caffeine that will be extracted is

$P = m * [\frac{V}{K * v_c + V} ]^3$

substituting values

$P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3$

$P = 39.8 \ mg$

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4 years ago

The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, an

Andre45 [30]

The height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

<h3>What is Pascal? </h3>

Pascal is defined as the force per unit area. It expressed in Newton pr square meter of area.

1 Pa = 1 N / m²

Pressure = force / Area

According to the question, the expression of force is as given below

F = mg

where,

F is the force,

m is the mass,

g is equal to the acceleration due to the gravity

Now, the area of the atmosphere is 1 m².The pressure is 1 atm. Pressure in Pascal.

1 atm = 1.01325 × 10 5pa

Therefore, the expression for pascal become as follows.

1.01325 × 10 5 pa = mg /area

1.01325 × 10 5 pa = m × 9.81 m/s² / 1 m²

M = 1.01325 × 10 5 pa × 1 m² / 9.81 m² × 1 Nm -² /1 pa × 1 kg m-² / 1 N

1.03 × 10 ⁴ kg

Given,

The density is 22.6 g /mL , pressure is 1 atm, and area is 1 m²

The relation between density and pressure can be given as follows.

P = hpg… … …(1 )

were , h is the height of the column

p is the density.

Hpg = 1.01325 × 10 5 pa × 1 N/m² /1 pa

H = 1.01325 × 10 5 N/m² / pg × 1 kg ms-² / 1 N

= 1.01325 × 10 5 kg m-¹ s -² / 22.6 g mL -1 × 1 kg/ 10 ³ g × 1 mL / 10 -6 m ³ × 9.81 m s- ²

= 0.457 m

Therefore, the height of the column is 0.457 m.

Thus, we concluded that the height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

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2 years ago

Which of the following is the correct name for MgCI?

kupik [55]

Answer:

C - Magnesium Chloride

Explanation:

The metal in a compounds name is always pronounced, and if the last element is a non-metal, the first half of the element is used and the last part is replaced with "ide"

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