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bezimeni [28]
3 years ago
13

Carboxyhemoglobin may be formed when a person inhales air that contains

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0
Answer is none of the above
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If a car travels 400.0 meters in 20.0 seconds, how fast is it going?<br> m/s<br> Please help me
Viefleur [7K]

Answer:

20m/s.

Explanation:

8 0
3 years ago
What thype of bond requires the give and take of electrons ?
DedPeter [7]
Ionic bond involves electrostatic attraction between oppositely charged ions.
The ions are atoms that have gained 1 or more electrons and atoms that have lost 1 or more electrons.
<span>Answer: The type of bond that requires the give and take of electrons is Ionic bond</span>
6 0
4 years ago
Read 2 more answers
How many grams of solute are needed to prepare a 3.50% mass/mass solution that has a solution mass of 2.50x102 grams.
Elis [28]
One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.

Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
5 0
3 years ago
Consider the reaction below. At 500 K, the reaction is at equilibrium with the following concentrations. [PCI5]= 0.0095 M [PCI3]
Dmitriy789 [7]

Answer: The equilibrium constant for the given reaction is 0.0421.

Explanation:

PCl_5\rightleftharpoons PCl_3+Cl_2

Concentration of [PCl_5] =  0.0095 M

Concentration of [PCl_3] =  0.020 M

Concentration of [Cl_2] =  0.020 M

The expression of the equilibrium constant is given as:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}

K_c=0.0421 (An equilibrium constant is an unit less constant)

The equilibrium constant for the given reaction is 0.0421.

6 0
3 years ago
Read 2 more answers
How would you prepare 3.5 L of a 0.9M solution of KCl?
PolarNik [594]
V=3,5L\\&#10;Cm=0,9M\\&#10;M_{KCl}=74\frac{g}{mol}\\\\&#10;C_{m}=\frac{n}{V}\\\\&#10;n=\frac{m}{M}\\\\&#10;C_{m}=\frac{m}{MV} \ \ \ \Rightarrow  \ \ \ m=C_{m}MV\\\\&#10;m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark. 
4 0
3 years ago
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