All categories

3 years ago

. The kidneys are _______________ to the liver.

Chemistry

1 answer:

m_a_m_a [10]3 years ago

7 0

I’d say posterior because the right kidney is slightly above where the liver is

You might be interested in

Analyze the bonding and structure of water and isopropanol. determine the type of bonding of each; ionic or covalent, polar or n

Naily [24]

Answer:

Water is polar covalent while isoporopanol is covalent

Explanation:

Water contains polar covalent bonds. Water has a high dipole moment and thus dissolves ionic substances easily such as sodium chloride. Water is also a good electrolyte. Water is not volatile. Water molecules are held together by hydrogen bonds. Isopropanol is a covalent compound. It is non polar and suitable for dissolving non polar substances such as campohor. It is volatile but dissolves in water due the the -OH group present in the molecule.

8 0

4 years ago

2 what is soil profile

marta [7]

Answer:

soil profile is the vertical cross- section of the soil containing various layers called horizons. it helps us to examine the structure of soil.

4 0

3 years ago

How many stereoisomers are obtained from the reaction of 3-methyl-2-butanone with each of the following organometallic reagents

nevsk [136]

Hey there!

Both are examples of Nucleophilic addition reaction and gives tertiary alcohol.

1 forms 2 Enantiomers namely (3S)-2,3-dimethyl-3-pentanol and its Enantiomer.

2 forms 2,3-dimethyl-2-butanol

3 0

3 years ago

Name this compound Ca (OH)2

Ira Lisetskai [31]

I believe the answer is Calcium dihydroxide

7 0

3 years ago

Read 2 more answers

Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final

Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

Step 1: Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

Step 2: The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

Step 3: Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

Step 4: Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

Step 5: Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

Step 6: Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0

3 years ago