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madam [21]
3 years ago
10

This is my cookie >:( you cannot have it

Chemistry
1 answer:
Fofino [41]3 years ago
6 0
Hmm, welll I think it’s heterogeneous
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Some states authorize a to make healthcare decisions ------- for patients who can't make them because they're incapacitated. O a
Tom [10]

Explanation:

A student flips four different coins at the same time. Which choice shows all of the possibilities for the results of the coin toss? (H = heads, T = tails)

A HH, TT, HT, TH

B HHHT, HTTT, HHTT

C HHH, TTT, HTH, HTT

D HHHH, HTHH, HTTH, HTTT, TTTT

7 0
3 years ago
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not
Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 =  [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

7 0
3 years ago
Guanidin HNC(NH2)2 is a fertilizer. What is the percent by mass of nitrogen in the fertilizer
Katyanochek1 [597]

Answer:

71.1%. , is a fertilizer.

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3 years ago
Why do ocean waves make a good source of energy?
JulsSmile [24]

Answer:

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6 0
3 years ago
If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch
Evgesh-ka [11]

Answer:

0.120M is the concentration of the solution

Explanation:

<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>

<em />

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

<em>Molar concentration:</em>

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>
7 0
3 years ago
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