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3 years ago

Write the molecular equation for sodium nitrate and lead (II) acetate

Chemistry

1 answer:

irga5000 [103]3 years ago

5 0

NaNO3 + Pb(C2H3O2)<span>2 would be the answer if you were adding the two. </span>

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This section of the periodic table is called a(n)

Alex_Xolod [135]

I’d say A yeah I’m pretty sure

3 0

3 years ago

Read 2 more answers

how much sodium chloride must be added to 100 mL of water so that its concentration is 20 parts per million assume that the dens

OlgaM077 [116]

Answer: 2mg

Part per million is unit that equal to mg/kg of water or simply $10^{-6}$ . In this case, you asked how much NaCl needed and know the volume of water(100mL), concentration(20ppm or 20 mg/kg) and water density.
Then, the equation would be

concentration = NaCl weight/ water weight
20 x $10^{-6}$ mg= NaCl weight/ (100ml x 1g/ml)
NaCl weight= 20x $10^{-6}$ mg/kg x 100g
NaCl weight= 2000 x $10^{-6}$ g
NaCl weight= 2x $10^{-3}$ g = 2mg

6 0

3 years ago

Choose the element that IS NOT in the same period as Potassium.

Lesechka [4]

E. sodium is the answer

8 0

4 years ago

Carboxypeptidases can be used to determine the entire sequence of short peptides (<50 amino acid residues) and it starts by r

zepelin [54]

Answer:

A. True

Explanation:

Carboxypeptidases are enzymes which function in the digestion of short peptides known as oligopeptides in the small intestine. Oligopeptides contain between 10 to 50 amino acid residues.

The two carboxypeptidases A and B involved in the digestion of proteins in the small intestine are secreted by the exocrine glands of the pancreas.They are both zinc-containing enzymes which remove successive carboxyl-terminal (C-terminal) residues from oligopeptides until free amino acids are obtained.

Since they function in this way, they can be used to determine the entire sequence of short peptides or oligopeptides.

6 0

3 years ago

How many milliliters of a 1M nitric acid solution are required to prepare 60mL of 6.7M solution?

Free_Kalibri [48]

Answer:

the number of milliliters of a 1M is 402mL

Explanation:

The computation of the number of milliliters could be determined by using the following formula

As we know that

$V_1\times M_1 = V_2\times M_2$

where,

V_1 and V_2 are the starting and final volumes

And, the M_1 and M_2 are the starting and the final molarities

Now the V_1 is

$V_1 \times 1M = 60mL \times 6.7M$

So, the V_1 is 402mL

Hence, the number of milliliters of a 1M is 402mL

4 0

3 years ago