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schepotkina [342]
3 years ago
13

I don’t understand this at all, if someone could help me please

Chemistry
1 answer:
rewona [7]3 years ago
6 0

Ok,

So what don't you understand. The whole page

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Which of the following equations has the correct products and is balanced correctly for a reaction between Na3PO4 and KOH?
ASHA 777 [7]

The answer is: A) Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.  

Na₃PO₄ is sodium phosphate.  

KOH is potassium hydroxide.  

NaOH is sodium hydroxide.  

K₃PO₄ is potassium phosphate.  

According to the mass conservation law, there are same number of atoms on both side of balanced chemical reaction.

5 0
3 years ago
Giving brainiest to whoever answers correctly.
Digiron [165]

Answer:250

Explanation:

5 0
3 years ago
What items could scientists use to date a layer of rock
Charra [1.4K]

By Using relative and radiometric dating methods hope this helps!!

4 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
An aqueous antifreeze solution is 60.0% ethylene glycol (HOCH2CH2OH) by mass and has a density of 1.06 g/mL. Calculate the molal
galina1969 [7]

Answer:

[HOCH₂CH₂OH] = 24.1 m

Explanation:

Ethylene glycol → HOCH₂CH₂OH

60% by mass means that 60 g of ethylene glycol are contained in 100 g of solution.

Solution mass = Solute mass + Solvent mass

100 g = 60 g + Solvent mass

Solvent mass = 40 g

Molality are the moles of solute contained in 1kg of solvent.

We determine the moles of solute → 60 g . 1mol/62 g = 0.967 moles

We convert the mass of solvent from g to kg → 40 g . 1kg/1000 g = 0.04 kg

Molality → 0.967 mol / 0.04 kg = 24.1 m

4 0
3 years ago
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