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2 years ago

9

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici

ents are:
ammonia (g) +
oxygen (g)
nitrogen monoxide (g) +
water (g)

1 answer:

tatuchka [14]2 years ago

5 0

The values of the coefficients would be 4, 5, 4, and 6 respectively.

<h3>Balancing chemical equations</h3>

The equation of the reaction can be represented by the following chemical equation:

ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)

$4NH_3(g)$ + $5O_2(g)$ ---> $4NO(g)$ + $6H_2O(g)$

Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.

More on balancing chemical equations can be found here: brainly.com/question/15052184

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

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The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

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In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

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$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

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$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

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$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

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