There are 2071.4662 grams of glucose in 11.5 moles.

Per 1 mole there are 180.15588 grams of glucose. 180.5588 x 11.5 =2076.4262

**Answer:**

The mass is 20, 89 grams of He.

**Explanation:**

We calculate the weight of 1 mol of He, from the atomic weight of each element obtained from the periodic table: Weight 1 mol He= 4, 0020602 g. 1 mol He-----4, 0020602 g. 5,22 mol He---x= (5,22 mol He x4, 0020602 g)/ 1 mol He= 20, 89 g

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**Answer:** 850.0 g/min.

**Explanation:**

- The rate of the reaction =
** (ΔC/Δt) **where,

**ΔC** is the change in concentration of reactants or products.

**Δt** is the change in time of the reaction proceeding.

- The rate is needed to be calculated in
** (g/min)**.

- We need to calculate
**the amount of the product in (g)** via using the relation **(n = mass / molar mass).**

- mass (g) = n x molar mass,

- n = 1.5 moles and molar mass of P₄O₁₀ = 283.88 g/mol.

**m** = 1.5 x 283.88 = **425.82 g**.

**ΔC = 425.82 g and Δt = 30 s / 60 = 0.5 min**.

**The rate of the reaction** = ΔC / Δt = (425.82 g / 0.5 min) = **851.64 g/min**.

**<em>can be approximated to 850.0 g/min.</em>**

Put simply, generators convert kinetic energy, which is based on movement, into electric energy. However, there are a number of different ways that this kinetic energy can be achieved. Most commonly, this electrical generation is created by using electromagnetic induction and by harnessing mechanical energy that causes a generator to rotate. Therefore, one of a generator’s most principal operations is the creation of kinetic energy.

**Answer:**

41.6 moles

**Explanation:**

We apply the Ideal Gases Law to solve this:

P . V = n . R . T

Volume is 1m³ but we need to convert to dm³ (L), because the units of R

1m³ . 1000 dm³/1m³ = 1000 dm³ → 1000L

Now we replace → 1 atm . 1000 L = n . 0.082L.atm/mol.K . 293.15K

(1 atm . 1000 L) / (0.082L.atm/mol.K . 293.15K) = n → 41.6 moles