Answer:
0.0847M is molarity of sodium hydrogen citrate in the solution
Explanation:
The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.
<em>Moles Na2C6H6O7:</em>
<em>Molar Mass:</em>
2Na: 2*22.99g/mol: 45.98g/mol
6C: 6*12.01g/mol: 72.01g/mol
6H: 6*1.008g/mol: 6.048g/mol
7O: 7*16g/mol: 112g/mol
45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol
Moles of 2g:
2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>
<em />
<em>Liters solution:</em>
100mL * (1L / 1000mL) = <em>0.100L</em>
<em>Molarity:</em>
8.473x10⁻³ moles / 0.100L =
<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
Answer:
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Explanation:
The empirical formula : Na₂Cr₂O₇
<h3>Further explanation</h3>
Given
170 g sample contains :
29.84 g sodium, 67.49 g chromium, and 72.67 g oxygen
Required
The compound's empirical formula
Solution
mol ratio of elements :
Na : 29.84 : 23 g/mol = 1.297
Cr : 67.49 : 51,9961 g/mol = 1.297
O : 72.67 : 16 g/mol = 4.54
Divide by 1.297
Na : Cr : O = 1 : 1 : 3.5 = 2 : 2 : 7
They are on the left side of the equation.
The balanced reaction is:
<span>Cu + 2AgNO3 = 2Ag + Cu(NO3)2
We are givent the amount of silver to be produced. This will be the starting point. Calculations are as follows:
</span><span> 3.50 mol Ag ( 1 mol Cu / 2 mol Ag ) = 1.75 mol Cu
</span>
Therefore, we need 1.75 mol of copper in order to produce 3.50 mol of silver.
