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Fiesta28 [93]
3 years ago
6

I have questions

Chemistry
1 answer:
marishachu [46]3 years ago
5 0
A) Al(NO3)2
b)CaCO3
c)AlPO4
d)NH4NO3
e)Ba3(PO)4
Hope this helps you;)
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Summarize the Natural Processes of weathering ,erosion, and deposition
Lena [83]

Answer:

I won’t do it in paragraph form cuz it will look very choppy but here you go:

Weathering is when the weather itself changes something, like when a metal bike gets rusty after sitting outside for a long time, or when a plant grows out of concrete.

Erosion is when something gets eroded away at. Like when something has water or wind is flowing against it so much that it changes shape. This is how canyons are made.

Deposition is when a gas turns into a solid, removing energy and skipping the liquid step (frost forming on car window)

7 0
3 years ago
Read 2 more answers
Now, examine the structures of benzhydrol and fluorene. Both compounds contain the same number of carbons but have very differen
zubka84 [21]

Answer:

The OH group

Explanation:

Benzhydrol contains OH hydroxyl group in its molecule while fluorene does not. At first glance, one would think that OH, which contributes to hydrogen bonding would causes melting point of benzhydrol to be higher than fluorene. <em>However, </em>the structure of benzhydrol, which is 2 benzene rings connected to center hydroxyl carbon (PhCOHPh), allows for each benzene rings in benzhydrol to rotate until both rings are perpendicular to minimize repulsive force. This prevents the molecule from stacking on each other due to its non flat shape, and thus, lowering its melting point in contrast to flat fluorene molecule.

4 0
3 years ago
There is only one correct way to conduct a scientific investigation. Is this true or false?
MA_775_DIABLO [31]
FALSE there are many ways to conduct scientific investigations 
7 0
3 years ago
How many grams of lead ii sulfide is produced when 25.0 g lead ii acetate reacts with excess hydrogen sulfide
ddd [48]

Answer: 18.42 grams of lead (II) sulfide will be produced in the given reaction:

Explanation: The reaction of lead (II) acetate and hydrogen sulfide follows:

(CH_3COO)_2Pb+H_2S\rightarrow PbS+2CH_3COOH

To calculate the moles, we use the formula:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}    ....(1)

Molar mass of lead (II) acetate = 325.29 g/mol

Given mass of lead (II) acetate = 25 g

Putting values in above equation, we get:

Moles=\frac{25g}{325.29g/mol}=0.0768moles

We are given that hydrogen sulfide is present in excess, so limiting reagent is lead (II) acetate because it limits the formation of product.

By stoichiometry of the reaction,

1 moles of lead (II) acetate produces 1 mole of lead (II) sulfide

So, 0.0768 moles of lead (II) acetate will produce = \frac{1}{1}\times 0.0768 = 0.0768 moles of lead (II) sulfide.

Now, to calculate the mass of lead (II) sulfide, we use equation 1, we get:

Molar mass of lead (II) sulfide = 239.3 g/mol

0.0768mol=\frac{\text{Given mass}}{239.3g/mol}

\text{Mass of lead (II) sulfide}=18.42g

8 0
3 years ago
Use the problem below to answer the question: When a current is passed through a solution of salt water, sodium chloride decompo
allochka39001 [22]
M(NaCl)/{2M(NaCl)}=m(Cl₂)/M(Cl₂)

m(Cl₂)=M(Cl₂)m(NaCl)/{2M(NaCl)}

m(Cl₂)=70.90*7.5/{2*58.44}=4.55 g

m(Cl₂)=4.55 g
3 0
3 years ago
Read 2 more answers
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