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Maru [420]
3 years ago
11

ASAP MULTIPLE CHOICE WILL MARK BRAINLIEST

Chemistry
1 answer:
yaroslaw [1]3 years ago
6 0

Answer:

There are four electrons

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How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?
zalisa [80]

Answer:

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl

The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g × (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 × 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol × (126.75 g/mol) = 47.2 g

5 0
3 years ago
Please help with this question.
marysya [2.9K]

Answer:

It is C.

Explanation:

4 0
3 years ago
What is the molar mass of Agl2​
Maru [420]

Answer:

107.8682

Explanation:

....................

7 0
3 years ago
Which of the following statements is typically true for a catalyst? I. The concentration of the catalyst will go down as a react
gladu [14]

Answer: II. The catalyst provides a new pathway in the reaction mechanism.

III. The catalyst speeds up the reaction.

Explanation:

Activation energy is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more molecules can cross the energy barrier and more molecules convert to products.

The catalyst itself does not take part in the chemical reaction and gets regenerated as such at the end of the reaction without getting consumed.

3 0
3 years ago
Now let us refine our model by noting that there is a second source of Na+ ions in the cell: NaI. Suppose the outside of the cel
MArishka [77]

Answer:

E=55mV

Explanation:

Hello,

Considering the given information, the new concentration of Na+inside will be (13.6 + 4) mM = 17.6 mM, and outside will be (150 + 0.04) mM = 150.04 mM due to the previous specified conditions.

Now, the recalculation of the Nerst potential at the supposed temperature of 25 °C (which is modifiable) is done via:

E=\frac{RT}{zF} ln(\frac{C_{Na^+,outside}}{C_{Na^+,outside}} )\\\\E=\frac{8.314\frac{J}{mol*K}*298K}{1*9.65x10^4\frac{C}{mol} } *ln(\frac{150.04}{17.6} )=0.055V*\frac{1x10^3mV}{1V} \\E=55mV

Best regards.

8 0
3 years ago
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