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ololo11 [35]
3 years ago
15

A student prepares a solution of potassium chloride by dissolving 90.6 g of KCI into

Chemistry
1 answer:
Wittaler [7]3 years ago
8 0

Answer:

B

Explanation:

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A
vodka [1.7K]
In my opinion the answer is identical
3 0
3 years ago
Read 2 more answers
The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation
Reil [10]

Answer:

1027.62 g

Explanation:

For B_2H_6  :-

Mass of B_2H_6  = 296.1 g

Molar mass of B_2H_6  = 27.66 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{296.1\ g}{27.66\ g/mol}

Moles\ of\ B_2H_6= 10.705\ mol

From the balanced reaction:-

B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}

1 mole of B_2H_6 react with 3 moles of oxygen

Thus,

10.705 mole of B_2H_6 react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

8 0
3 years ago
The function of the heat exchanger in a nuclear reactor is to...
Ganezh [65]

Answer:

cool the tractor water system

4 0
2 years ago
Calculate the amount of water required to prepare 500g of 2.5% solution of sugar.
Snowcat [4.5K]

(i) We start by calculating the mass of sugar in the solution:

mass of sugar = concentration × solution mass

mass of sugar = 2.5/100 × 500 = 12.5 g  

Then now we can calculate the amount of water:

solution mass = mass of sugar + mass of water

mass of water =  solution mass - mass of sugar

mass of water = 500 - 12.5 = 487.5 g

(ii) We use the following reasoning:

If       500 g solution contains 12.5 g sugar

Then    X g solution contains 75 g sugar

X=(500×75)/12.5 = 3000 g solution

Now to get the amount of solution in liters we use density (we assume that is equal to 1):

Density = mass / volume

Volume = mass / density

Volume = 3000 / 1 = 3000 liters of sugar solution

8 0
3 years ago
Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0
3 years ago
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