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2 years ago

What electrolyte may fall in level as a result of citrate toxicity from massive transfusion quizlket

1 answer:

7 0

Potassium Recipients of massive transfusions may therefore develop electrolyte disturbances, with hypocalcemia, hypomagnesemia, and hyperkalemia most commonly reporte

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Help it's science so help

bija089 [108]

1.b
2.b
3.a
3 was the the opposite of the scenario stated

7 0

3 years ago

Physical properties are :

elena-14-01-66 [18.8K]

A. properties which can be observed since it's what you're physically viewing

5 0

3 years ago

Which of the following models represents an atom that is more reactive than the

Zanzabum

Answer:

Model A

Explanation:

Model A represents an atom that is more reactive than the others represented.

Valence electrons actually determine the reactivity of elements. They also determine the properties of elements.

Elements with one valence electron are highly reactive because they need low energy to remove them. They can either gain more electrons to become stable or they share/give out their electrons.

Therefore, Model A is the correct answer because it has one valence electron and its valence electron is farther from the nucleus thereby this makes it more reactive.

8 0

3 years ago

Calculate the standard cell potential of the following cell at 25°C. Sn(s) | Sn2+(aq) || Cu2+(aq) | Cu(s).

Semmy [17]

Answer:

0.48 V

Explanation:

Usually in the cell notation, the left side shows oxidation. So,

Oxidation half reaction:

$Sn_{(s)}\rightarrow Sn^{2+}_{(aq)}+2e^-$ $E^o_{ox}=-E^o_{red}=-(-0.14\ V)=0.14\ V$

Reduction half reaction:

$Cu^{2+}_{(aq)}+2e^-\rightarrow Cu_{(s)}$ $E^o_{red}=0.34\ V$

$E^o_{cell}=E^o_{ox}+E^o_{red}= (0.14+0.34)\ V=0.48\ V$

4 0

3 years ago

A company is testing drinking water and wants to ensure that ca content is below 155 ppm. What is the maximum amount of Ca that

BabaBlast [244]

Answer:

138 mg

Explanation:

A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.

Step 1: Convert the mass of drinking water to kilograms.

We will use the relation 1 kg = 1000 g.

$890g \times \frac{1kg}{1000g} =0.890kg$

Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water

$0.890gH_2O \times \frac{155mgCa}{1kgH_2O} = 138mgCa$

7 0

3 years ago