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2 years ago

10

To produce 8 molecules of ammonia, NH3, we would need _ molecules of hydrogen, H2 and __ molecules of nitrog

en, N2

1 answer:

Setler79 [48]2 years ago

7 0

I have included the work and the process which is easiest for me to balance chemical equations. I hope it helps!

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If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

2 years ago

What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?

White raven [17]

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

Temperature (T) = 353 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

1 × V = 1.27 × 0.0821 × 353

V = 36.8 L

Thus, the volume occupied by the helium gas is 36.8 L

5 0

3 years ago

How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?

Bad White [126]

The balanced chemical reaction is :

$O_2 + 4Na \ -> \ 2Na_2O$

Number of moles of Na, $n = \dfrac{14.6}{23} = 0.635 \ mol$ .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

$n = \dfrac{0.635}{4}\ mole$

So, amount of oxygen required is :

$m = \dfrac{0.635 \times 32}{4}\ gm\\\\m = 5.08 \ gm$

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0

2 years ago

Balance the equation by ion electron method

irina1246 [14]

Answer:i believe you are to decompose the formula (i think)

8 0

2 years ago

Is the flame coloration a test for the metal or for nitrate ion?

mel-nik [20]

Metallic ion because the metallic ions will enter an exited state and release photons energy, in the form of light, as they return to their ground state.

5 0

3 years ago

Read 2 more answers