Round the number 3456.5 to two significant figures.

When the procedure calls for making a more dilute solution of an acid, or mixing an acid with other solutions, what is the corre

stiks02 [169]

Explanation:

Whenever we need to make a dilute solution of an acid then it is necessary to add water or non-acidic component into the acid first. This is because addition of water or non-acidic component directly into the acid could be highly exothermic in nature.

As a result, the acid can splutter and can cause burning of skin and other serious damage.

So, in order to avoid such type of damage the addition of water or non-acidic component into the acid actually helps to minimize the heat generated.

Thus, we can conclude that correct order of steps for making a more dilute solution of an acid is that either add all of the water or non-acid component first, or add a significant portion, before adding the acid to the mixture.

8 0

3 years ago

Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of

ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

P= 1.10 atm
V= 26.5 L
n=?

$R=0.082057 \frac{atm*L}{mol*K}$

T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* $0.082057 \frac{atm*L}{mol*K}$ *295 K

Solving:

$n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}$

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

$moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }$

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

$mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}$

mass of NaN₃=52.008 grams

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

6 0

3 years ago

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3. What is the mass of 5 moles of Hydrogen sulfate?I

solong [7]

Answer:

490

98 for 1 mole, Hence for 5 moles 5 X 98 =490.

Explanation:

Brainliest please?

3 0

3 years ago

34. The line shows the path of the ball.

kiruha [24]

C. Potential to kinetic

6 0

3 years ago

Read 2 more answers

An acid

Arte-miy333 [17]

An acid a. has a high pH in solution.
b. turns blue litmus paper to red.
c. releases hydroxyl ions in solution.
d. has more hydroxyl than hydrogen (or hydronium) ions.

An acid turns blue litmus paper to red. The answer is letter B.

4 0

3 years ago