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qaws [65]
3 years ago
12

What Type of chemical reaction is: BaCl2 + 2 AgNO3 --> 2 AgCl + Ba(NO3)2

Chemistry
1 answer:
romanna [79]3 years ago
8 0

Answer:

This is a precipitation reaction: AgCl is the formed precipitate.

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How many grams of hydrogen are in 46 g of CH40?<br>​
Romashka [77]

Answer:

5.8 grams of hydrogen

Explanation:

Your welcome

6 0
2 years ago
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The compound, nah2po4, is present in many baking powders. what is its name? sodium phosphate sodium bephosphate sodium dihydroge
AlladinOne [14]
 The  compound NaH2PO4     name  is

sodium  dihydrogen  phosphate

 
Explanation
This name is    arrived  at  by using the   IUPAC rules  of naming   compound

1.  the metal (sodium)is  named  first  followed  by  the ligand ( hydrogen  and phosphate)

Ligand  are  molecules   that are   attached to  the metal  center.

2.   ligand  are  named using alphabetical   order(for  our  case  h  for hydrogen come before p for  phosphate hence hydrogen is named first)

3. Prefix   di is used  since hydrogen  are two

hence the name  of the compound  is  Sodium dihydrogen  phosphate


8 0
3 years ago
What is H20 and oxygen
olga2289 [7]
H2O means water, which is made of two hydrogen and one oxygen atoms.
Oxygen is a gas that your body needs to function. Oxygen comes from the air.
4 0
3 years ago
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Describe how sediments are produced by weathering and erosion
TEA [102]
Weathering is the physical or chemical process where rocks are broken down to form sediments.
 • Physical weathering does not change the rock’s composition.
 • Chemical weathering does change the rock’s composition
 • Erosion is the process where forces such as water, wind, ice, or gravity transport sediments to some remote location
5 0
3 years ago
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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
  • brainly.com/question/19671384
5 0
2 years ago
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