Answer:
1.18 moles of CS₂ are produced by the reaction.
Explanation:
We present the reaction:
5C + 2SO₂ → CS₂ + 4CO
5 moles of carbon react to 2 moles of sulfur dioxide in order to produce 1 mol of carbon disulfide and 4 moles of carbon monoxide.
As we do not have data from the SO₂, we assume this as the excess reagent. We convert the mass of carbon to moles:
70.8 g / 12 g/mol = 5.9 moles
Ratio is 5:1, so 5 moles of carbon react to produce 1 mol of CS₂
Then, 5.9 moles will produce (5.9 . 1) / 5 = 1.18 moles
An isotope is defined as an element that has the same number of protons as the common element but different number of neutrons. In this case, a beryllium atom has a molar weight of 10 amu. Thus, there are 4 protons and 6 neutrons. The nuclear symbol of Be-10 is 4 Be10
Explanation:
<em>2</em><em>1</em><em> </em><em>piec</em><em>e</em><em> </em><em>=</em><em>4</em><em>5</em><em> </em><em>mins</em>
<em>100</em><em> </em><em>pieces</em><em>=</em><em> </em><em>x</em>
<em>on</em><em> </em><em>cross</em><em> </em><em>mul</em><em>tiplicati</em><em>on</em>
<em>2</em><em>1</em><em>x</em><em>=</em><em>4</em><em>5</em><em>×</em><em>1</em><em>0</em><em>0</em><em>m</em><em>i</em><em>n</em><em>s</em>
<em>2</em><em>1</em><em>x</em><em> </em><em>=</em><em>4</em><em>5</em><em>0</em><em>0</em><em>m</em><em>i</em><em>n</em><em>s</em>
<em>x</em><em> </em><em>=</em><em>4</em><em>5</em><em>0</em><em>0</em><em>m</em><em>i</em><em>n</em><em>s</em><em>÷</em><em>2</em><em>1</em>
<em> </em><em> </em><em>=</em><em>2</em><em>1</em><em>4</em><em>.</em><em>3</em><em>m</em><em>i</em><em>n</em><em>s</em><em> </em><em>or</em><em> </em><em> </em><em>3</em><em>.</em><em>5</em><em>7</em><em>h</em><em>r</em><em>s</em>
Question:
A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Answer:
The answer to the question is as follows
The mass of solution the student should use is 23.42 g.
Explanation:
To solve the question we note the following
A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution
Therefore we have 10 g of isopropenylbenzene contained in
100 g * 10 g/ 42.7 g = 23.42 g of solution
Available solution = 120 g
Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.
