Nitric acid, , is manufactured by the Ostwald process, in which nitrogen dioxide, , reacts with water. How many grams of nitrogen dioxide are required in this reaction to produce 6.40 g ? g

7.04 g

Explanation:

Let's consider the reaction in the last step of the Ostwald process.

3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:

6.40 g × (1 mol/63.01 g) = 0.102 mol

The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:

0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂

The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:

0.153 mol × (46.01 g/mol) = 7.04 g

an honest friend. a friend who cares about me, and wont use me. a friend that i can trust. not someone who is fake

mark this brainliest lol i need one more.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml