Which of the following are elements in this equation:

A solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet

Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

$m=1.67 g/mL\times 29.1 mL=48.597 g$

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= $\frac{70.5}{100}\times 48.597 = 34.261 g$

Moles of perchloric acid = $\frac{34.261 g}{100.46 g/mol}=0.3410 mol$

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

$Molarity=\frac{Moles}{Volume (L)}$

$=\frac{0.3410 mol}{0.250 L}=1.364 M$

The concentration of the solution is 1.364 molar.

6 0

3 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

3 years ago

How many moles are present in 63.80L of oxygen gas, O2?

matrenka [14]

Answer:

2.85moles of oxygen gas

Explanation:

Given parameters:

Volume of oxygen gas = 63.8L

Unknown:

Number of moles = ?

Solution:

We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:

1 mole of a gas at STP occupies a volume of 22.4L

So;

63.8L of oxygen gas will take up a volume of

$\frac{63.8}{22.4}$ = 2.85moles of oxygen gas

4 0

3 years ago

Fill in the Blank. 1. You want to be shielded from all three types of nuclear radiation. If you find shielding that blocks _____

Lena [83]

1. If you find shielding that blocks gamma radiation,then it will most likely also block the other two types. (as the hardest radiation)
2. When nuclear radiation enters the tube of a Geiger counter, it the excited atoms of the gas contained in the tube. (this type <span>deposits all energy into a single atom)</span>
3 The particles that make up protons and neutrons and are thought to be basic units of matter are <span>quarks.</span>

6 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

shepuryov [24]

The Outer Core Is Made Up Of Molten Nickel And Iron

6 0

3 years ago