GIVE ME BRAINLIEST!!!! nah im jk lol ANYWAY HAVE A GREAT DAY!

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KonstantinChe [14]

Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

$1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L$

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given $BaSO_{4}$ solution is as follows.

$Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g$

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

7 0

3 years ago

If 0.138g of cyclohexene (c6h10) was obtained from 0.240g of cyclohexanol (c6h120), what is the percentage yield of cyclohexene?

tino4ka555 [31]

Given in the question- 1 mole of cyclohexanol = > 1 mole of cyclohexene Molar mass 100.16 g/mol moles of cyclohexanol = .240 / 100.16= 0.002396 moles Molar mass 82.143 g/mol moles of cyclohexene formed @100 % yield = 0.002396 Molar mass 82.143 g/mol mass of cyclohexene @ 100 % = .002396 x 82.143 = 0.197g bur we have .138g so % yield = .138 / .197 = 70.0 % Ans- 70 percentage yield of cyclohexene.

4 0

3 years ago

What type of soil holds the least amount of water

jeka57 [31]

The clay soil holds the least amount if water.

8 0

3 years ago

Read 2 more answers

What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?

Alika [10]

Answer:- B: $1.10g H_2$ is the right answer.

Solution:- The balanced equation is:

$O_2(g)+2H_2(g)\rightarrow 2H_2O(g)$

We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.

From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.

So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.

The complete set up looks as:

$8.75g O_2(\frac{1mole}{32g})(\frac{2mole H_2}{1mole O_2})(\frac{2.02g}{1mole})$