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2 years ago

15

You chill your reaction in an ice bath when adding nitric acid to the sulfuric acid. why?

1 answer:

Andrej [43]2 years ago

4 0

The reaction of nitric acid and sulfuric acid is highly exothermic so it releases a lot of heat. If the temperature is not controlled, the reaction could go into thermal runaway, which is potentially extremely hazardous.

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Q 9 The table lists the steps to clean water for drinking purpose. 1. Adding chlorine tablets to the water.2. Pouring the water

love history [14]

Answer:

1 and 2

Explanation:

when we r adding chlorine to water

the water is clean

8 0

2 years ago

Which property of a liquid increases with increasing temperature?

Nina [5.8K]

When temperature of liquid is increased, liquid gets thinner and thinner and hence it's viscosity decreases.

Density = mass/volume.

As we increase the temperature, volume of the liquid starts to increase but mass of the liquid remains constant. As a result, density of liquid decreases.

Hope this helps!

5 0

2 years ago

Answer quickly i'm out of time !!!!!!

melamori03 [73]

Answer:

Option D. pH= 1.3 strong acid

Explanation:

From the question given:

The hydrogen ion concentration [H+] = 0.05 M

pH = —Log [H+]

pH = —Log 0.05

pH = 1.3

Since the pH lies between 0 and 7, the solution is acidic. Since the pH value is low, the solution is a strong acid

8 0

2 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

What is the value of R in the ideal gas law?

dybincka [34]

It’s option B 0.0821L.atm/mol.K

5 0

2 years ago