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3 years ago

What happens after coral is bleached?

Chemistry

1 answer:

seropon [69]3 years ago

5 0

It releases algae into the ocean

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Which electron configuration is impossible?

Verizon [17]

Answer:

MP hippopotamus

Explanation: I'm in the fifth grade so yeah I really don't know this so I'm just going to say some random stuff and white cheddar mac and cheese warm TV perfect back MD 11443 to 2885 eleven 12:20 to 11 132 0.24 answer

7 0

3 years ago

A hydrogen halide diffuses 1.49 times faster than HBr. This hydrogen halide is

marysya [2.9K]

To solve this problem, we must assume ideal gas behaviour so that we can use Graham’s law:

vA / vB = sqrt (MW_B / MW_A)

where,

vA = speed of diffusion of A (HBR)

vB = speed of diffusion of B (unknown)

MW_B = molecular weight of B (unkown)

MW_A = molar weight of HBr = 80.91 amu

We know from the given that:

vA / vB = 1 / 1.49

So,

1/1.49 = sqrt (MW_B / 80.91)

MW_B = 36.44 g/mol

Since this unknown is also hydrogen halide, therefore this must be in the form of HX.

HX = 36.44 g/mol , therefore:

x = 35.44 g/mol

From the Periodic Table, Chlorine (Cl) has a molar mass of 35.44 g/mol. Therefore the hydrogen halide is:

HCl

6 0

3 years ago

Hellppppppppplplllppp

noname [10]

Answer:

the third one because oceans don't affect the weather

Explanation:

7 0

3 years ago

Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).

dedylja [7]

The balanced chemical equation between HCl and $Ba(OH)_{2}$ is:

$2HCl (aq) + Ba(OH)_{2}(aq) -->BaCl_{2}(aq) + 2 H_{2}O(l)$

Moles of $Ba(OH)_{2}$ = $1.00 g Ba(OH)_{2} * \frac{1 mol Ba(OH)_{2}}{171.3 g Ba(OH)_{2}} = 0.00584 mol Ba(OH)_{2}$

Moles of HCl required to neutralize $Ba(OH)_{2}$ :

$0.00584 mol Ba(OH)_{2} * \frac{ 2 mol HCl}{1 mol Ba(OH)_{2}} = 0.01168 mol HCl$

Calculating the volume of HCl from moles and molarity:

$0.01168 mol HCl * \frac{1 L}{0.100 mol} * \frac{1000 mL}{1 L} = 116.8 mL$

7 0

3 years ago

Read 2 more answers

Write an equation that shows the reaction between acetic acid and triethylamine (CH3CH2)3N. Draw all non-bonding lone electron p

Anni [7]

Answer:

-) Acid-base reaction

-) Carboxylic acid, alcohol, alkene and ketone

Explanation:

For the reaction between acetic acid and triethylamine, we will have an acid-base reaction. Therefore a salt would be produced in this case an "ammonium quaternary salt". Also, we have to remember that on this reaction the acid is the acetic acid and the base is the triethylamine. See figure 1

For the second question, we have to check the structure of Prostaglandin E1 in which we have the functional groups:

1) Carboxylic acid

2) Alcohol

3) Alkene

4) Ketone

See figure 2.

I hope it helps!

8 0

3 years ago