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Marrrta [24]
2 years ago
9

If you switch to an all plant diet, what vitamin will you need to take as a supplement? vitamin A,vitamin B12, vitamin C, vitami

n D​
Chemistry
2 answers:
rusak2 [61]2 years ago
6 0

Answer:

Vitamin B12

Explanation:

pashok25 [27]2 years ago
6 0

Answer: B12

Explanation: Took the test

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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
Is a nickel coin magnetic
AURORKA [14]

Answer:

Yes

Explanation:

Because it was false because it has been the substance abuse permalink

7 0
3 years ago
Read 2 more answers
Given the equation I = Q/t, solve for t.
larisa86 [58]

Answer: t=\frac{Q}{I}

Explanation:

I=\frac{Q}{t}

Multiply by t on both sides.

t*I=\frac{Q}{t}*t

tI=Q

Now divide by I to isolate t.

\frac{tI}{I}=\frac{Q}{I}

t=\frac{Q}{I}

5 0
3 years ago
What would indicate that a physical change takes place when cooper is drawn into wire
amid [387]

Answer:

When a copper is drawn into wire the only change that occurs is change in its shape and size no change will take place into its composition that is the wires are still possessing the properties of copper metal. Thus, a physical change takes place when copper is drawn into wire.

4 0
3 years ago
Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the
Olin [163]

Firstly we need to determine the partial pressure of O2:

\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\  \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}

We will now use the Henry's Law equation to determine the solubility of the gas:

\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\  \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}

Answer: Solubility is 2.7x10^-3 M

6 0
1 year ago
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