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2 years ago

Identify the independent /manipulated variable

Chemistry

1 answer:

fiasKO [112]2 years ago

7 0

Answer:

independent- test variable

manipulated - outcome variable; if needed, there could be changes.

Explanation:

knowledge

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Jim, Jill, Robert, and Kim each run paper chromatography on an unknown aqueous mixture. Jim gets a red band and a blue band, Jil

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Answer:

Robert

Explanation:

There is not more than one colour

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2 years ago

. How many grams of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide? ____Cl

dlinn [17]

Answer: 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide.

Explanation:

The balanced chemical reaction is;

$Cl_2+MgBr_2\rightarrow MgCl_2+Br_2$

$Cl_2$ is the limiting reagent as it limits the formation of product and $MgBr_2$ is the excess reagent.

According to stoichiometry :

1 mole of $Cl_2$ produces = 1 mole of $MgCl_2$

Thus 2 moles of $Cl_2$ will produce= $\frac{1}{1}\times 2=2moles$ of $MgCl_2$

Mass of $MgCl_2=moles\times {\text {Molar mass}}=2moles\times 95g/mol=190g$

Thus 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide

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2 years ago

When water reacts with potassium metal the hydrogen produced ignites explosively on the surface of water .What cause the ignitio

Julli [10]

Answer:

Pottasium reacts with water vigorously and the reation is exothermic. The heat released causes the hydrogen released to ignite

Explanation:

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2 years ago

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

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2 years ago

The specific heat of aluminum is s 9.00 J/g C how much energy is needed to raise the temperature of 20.0 g block of aluminum fro

Nimfa-mama [501]

E=MC(delta)T
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2 years ago