**Answer:**

**0.6749 M is the concentration of B after 50 minutes.**

**Explanation:**

A → B

Half life of the reaction =

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

Initial concentration of A =

Final concentration of A after 50 minutes =

t = 50 minute

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

**0.6749 M is the concentration of B after 50 minutes.**

**Answer:**

group 17 the halogen.as it has 7 electron in its outermost ring

The **weighted average** of the **nail** in accordance with the given data is 11.176g.

<h3>How to calculate weighted average?</h3>

**Weighted average **is an arithmetic mean of values biased according to agreed weightings.

The **weighted average** of the nail in the image above can be calculated by multiplying the decimal abundance with the **mass** of the **nail**, then summed up as follows;

**Weighted average** = (decimal abundance × mass 1) + (decimal abundance × mass 2)

**Weighted average** = (0.12 × 3.3) + (0.88 × 12.25)

**Weighted average **= 0.396 + 10.78

**Weighted average** = 11.176g

Therefore, 11.176g is the **weighted average** of the **nail**

Learn more about **weighted average **at: brainly.com/question/28042295

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**Answer : Right**

**Explanation :** The direction of reaction tends to proceed on right side under standard conditions; If the change in standard free energy ΔG for a particular reaction is negative. Also if the elements in their most stable forms as they exist under standard conditions. Then ΔG determines the direction and extent of chemical change. But under standard conditions the direction of the reaction will be to right.

**Answer:**

**The concentration of the solution is 5.8168 × ** mol.

**Explanation:**

Here, we want to calculate the concentration of the solution.

The unit of this is mol/dm^3

So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3

The number of moles = mass/molar mass

molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol

Number of moles = 33.4/87 = 0.384 moles

This 0.384 moles is present in 660 L

x moles will be present in 1 dm^3

Recall 1 dm^3 = 1L

x * 660 = 0.384 * 1

x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3