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2 years ago

13

What information do the chemical hazard label and msds have in common

1 answer:

snow_tiger [21]2 years ago

6 0

They tell you to have caution.

You might be interested in

What do lightning and stars have in common?

irina1246 [14]

2... one though it is true is not very strong

three is very incorrect and dumb

four... idek

4 0

2 years ago

What is the density of a substance with a mass of 1.8 kg and a volume of 0.2 L?

horrorfan [7]

Answer:

<h3>The answer is 9.0 kg/L</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 1.8 kg

volume = 0.2 L

We have

$density = \frac{1.8}{0.2} \\$

We have the final answer as

<h3>9.0 kg/L</h3>

Hope this helps you

5 0

3 years ago

Explain why hydrogen will form an ionic bond with fluorine but only forms covalent bonds with carbon

vaieri [72.5K]

A fluorine atom has seven valence electrons. ... Carbon will then have five valence electrons (its four and the one its sharing with fluorine). Covalently sharing two electrons is also known as a “single bond.” Carbon will have to form four single bonds with four different fluorine atoms to fill its octet.

4 0

2 years ago

Read 2 more answers

Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc

marysya [2.9K]

Explanation:

1) $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)$

$Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091$

2) $H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)$

$Kc=\frac{[H_2SO_4]}{[SO_3]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3$

3) $P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)$

$Kc=\frac{1}{[O_2]^3}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641$

5 0

3 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

2 years ago