I need help on this 6th grader science plants.

Name an element or elements in the periodic table that you would expect to be chemically similar to potassium.

Rina8888 [55]

Potassium (K) is in Group I of the periodic table, and elements in the same column (period) are similar. Sodium (Na), or lithium (Li) are similar.

6 0

3 years ago

Help me, I don’t understand

marysya [2.9K]

Answer:

option c

Explanation:

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7 0

3 years ago

Read 2 more answers

Which best describes a similarity between power plants that use water as an energy source and those that use wind as an energy s

elixir [45]

Answer:

Both use kinetic energy to produce electricity.

6 0

3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago

A solution is prepared from 4.5701 g of magnesium chloride and 43.238 g of water. The vapor pressure of water above this solutio

balandron [24]

Answer:

i = 2.483

Explanation:

The vapour pressure lowering formula is:

Pₐ = Xₐ×P⁰ₐ (1)

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ

Where:

Pₐ is vapor pressure of solution (0.3624atm), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (0.3804atm)

4.5701g of MgCl₂ are:

4.5701g ₓ (1mol / 95.211g) = 0.048000 moles

43.238g of water are:

43.238g ₓ (1mol / 18.015g) = 2.400 moles

Replacing in (1):

0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm

0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)

2.4mol + i*0.048mol = 2.4mol / 0.9527

2.4mol + i*0.048mol = 2.5192mol

i*0.048mol = 2.5192mol - 2.4mol

i = 0.1192mol / 0.048mol

i = 2.483

I hope it helps!

4 0

3 years ago