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3 years ago

13

are medications typically used to relieve excessive mucus in nasal pathways. a) Radioactive isotopes b) Expectorants c) Deconges

tants O d) Antibiotics and anti-inflammatories

1 answer:

Mrac [35]3 years ago

8 0

Answer:

c) decongestants relive excessive muscle in nasal pathways

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What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte in

Vika [28.1K]

Explanation:

Apple cider vinegar or balsamic vinegar might contain various synthetics which might interact with or nullified the findings, whereas white vinegar comprises of acidic acid only. In fact, apple cider vinegar and balsamic vinegar are deep in appearance this would make it very difficult to determine the color. This is why it is preferable to use white vinegar in the Titration process.

3 0

3 years ago

Covalent bonded atoms make up what

White raven [17]

In a covalent bond, the atoms bond by sharing electrons. Covalent bonds usually occur between nonmetals. For example, in water (H2O) each hydrogen (H) and oxygen (O) share a pair of electrons to make a molecule of two hydrogen atoms single bonded to a single oxygen atom.

7 0

3 years ago

If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a

Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0

3 years ago

Please help with the question attached<br>I need it in an hour

lawyer [7]

Answer:

iiiiiiddddiiilllooeiuehe

4 0

2 years ago

Calculate the amount of energy required to melt 500 g of ice at 0oc. (δhfus=5.96 kj/mole; δhfus is the energy required to melt i

marta [7]

When you want to melt an ice, you only need the latent energy of fusion, <span>δhfus. We use the given value, then multiply this with the given amount to determine the amount of energy. Since the energy is per mole basis, use the molar mass of ice which is 18 g/mol. The solution is as follows:

</span>ΔH = 5.96 kJ/mol * 1 mol/18 g * 500 g
<em>ΔH = 165.56 kJ</em><span>
</span>

3 0

3 years ago