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3 years ago

When it comes to safe patient handling, what does it take to perform this task ergonomically

1 answer:

6 0

When it comes to safe patient handling, what it takes to perform this task ergonomically is commitment. This is a very essential training given to those who are handling patients in order to avoid musculoskeletal injuries and damages. Hope this answers your question. Have a great day ahead!

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Convert 779 gallons to cubic yards

AfilCa [17]

The answer would be About 3.857yd3

7 0

3 years ago

The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so th

LekaFEV [45]

Answer:

A penny dissolves in hydrochloric acid if the copper coating is filed down in several spot... ... When The Zinc In A Certain Penny Dissolves, The Total Volume Of Gas ... is filed down in several spots (so that the hydrochloric acid can reach the zinc). The reaction between the acid and the zinc 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq) .

6 0

3 years ago

What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

3 years ago

Which accurately completes the statement? Conduction involves the transfer of ___________ due to the movement of particles. Grou

Jlenok [28]

Explanation:

conduction involves the transfer of molecules due to the movement of particles

5 0

3 years ago

Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4

vova2212 [387]

Answer:

a) 226.6 grams of Cl₂

b) 19.2 grams of C

c) 303.2 grams of TiCl₄ and 70.4 grams of CO₂

Explanation:

The balanced chemical reaction is the following:

TiO₂(s) + C(s) + 2 Cl₂(g) → TiCl₄(s) + CO₂(g)

(a) What mass of Cl₂ gas is needed to react with 1.60 mol TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 2 moles of Cl₂. So, the stoichiometric ratio is 2 mol Cl₂/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of Cl₂ we need:

1.60 mol TiO₂ x 2 mol Cl₂/1 mol TiO₂ = 3.2 mol Cl₂

Now, we convert from moles to mass by using the molecular weight (MW) of Cl₂:

MW(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

mass of Cl₂= 3.2 mol x 70.8 g/mol = 226.6 g

Therefore, 226.6 grams of Cl₂ are needed to react with 1.6 mol of TiO₂.

(b) What mass of C is needed to react with 1.60 mol of TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 1 moles of C(s). So, the stoichiometric ratio is 1 mol C/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of C(s) we need:

1.60 mol TiO₂ x 1 mol C(s)/1 mol TiO₂ = 1.60 mol C(s)

So, we convert the moles of C(s) to grams as follows:

MW(C) = 12 g/mol

1.60 mol x 12 g/mol = 19.2 g C(s)

Therefore, a mass of 19.2 grams of C is needed to react with 1.60 mol of TiO₂.

(c) What is the mass of all the products formed by reaction with 1.60 mol of TiO₂?

From the chemical equation, we can notice that 1 mol of TiO₂ produces 1 mol of TiCl₄ and 1 mol of CO₂. So, from 1.60 moles of TiO₂, 1 mol of each product will be produced:

1 mol TiO₂/1 mol TiCl₄ ⇒ 1.60 mol TiO₂/1.60 mol TiCl₄

1 mol TiO₂/1 mol CO₂ ⇒ 1.60 mol TiO₂/1.60 mol CO₂

Finally, we convert the moles to grams by using the molecular weight of each compound:

MW(TiCl₄) = 47.9 g/mol Ti + (35.4 g/mol x 4 Cl) = 189.5 g/mol

1.60 mol x 189.5 g/mol = 303.2 g

MW(CO₂) = 12 g/mol C + (16 g/mol x 2 O) = 44 g/mol

1.60 mol x 44 g/mol = 70.4 g

Therefore, from the reaction of 1.60 mol of TiO₂ are formed 303.2 grams of TiCl₄ and 70.4 grams of CO₂.

3 0

3 years ago