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3 years ago

How do yo know when your in love ?

Chemistry

1 answer:

balu736 [363]3 years ago

8 0

Answer:

Play the 24 signs you're falling in love bingo I attached lol (I hope this isn't for school)

Explanation:

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What is the standard cell notation of a galvanic cell made with aluminum and maggnesium?

BartSMP [9]

Mg( s ) | Mg 2+ ( aq ) || Al 3+ ( aq ) | Al( s )

3 0

3 years ago

You are asked to calculate the concentration of a NaCl (in water) solution in several ways and are given the following informati

atroni [7]

B) The mass percentage of a solution is the ratio of the mass of solute to the mass of solvent expressed as a percentage. This becomes:
mass % = (NaCl mass) / (water mass) x 100
Mass % = 20 / 250 x 100
Mass % = 8%

C)
The molarity of a solution is the moles of solute present per liter of solution. The volume of the solution is:
250 mL = 0.25 L
Moles = mass / Mr
Moles = 20 / 58.44
Moles = 0.34

Molarity = moles / liter
Molarity = 0.34 / 0.25
Molarity = 1.36 M

3 0

3 years ago

water at 400 kPa and 60°C while the other part is evacuated. The partition is now removed, and the water expands to fill the ent

kaheart [24]

Answer:

0.4912 kJ/K

Explanation:

Using steam table, the properties of water at stage 1 can be determined:

$P_{1}$ = 400 kPa

$T_{1}$ = 60°C

Therefore, from steam table:

specific volume $v_{1} = 0.001017 m^{3}/kg$

entropy $s_{1} = 0.8313 kJ/kg*K$

Calculating the specific volume of water after the partition is removed, since the mass is constant and volume of water is doubled:

$v_{2} = 2*0.001017 = 0.002034 m^{3}/kg$

Final pressure of water in the tank = $P_2} = 40 kPa$

dryness factor after removing the partition

$x = \frac{0.002034 - 0.001026}{3.993 - 0.001026} = 2.524*10^{-4} kg/vapor$

Therefore:

$s_{2} = s_{f2} + x*s_{fg2} = 1.0261 + (2.524*10^{-4}*6.643) = 1.02778 kJ/kg*K$

Thus, the entropy change is:

Δs = $m*(s_{2}-s_{1}) = 2.5*(1.02778 - 0.8313) = 0.4912 kJ/K$

Therefore, the entropy change of water is 0.4912 kJ/K

6 0

4 years ago

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

HELP ASAP WILL GIVE BRAINIEST IF CORRECT a a a²+ b2 = c² ? What is the main benefit of this model? O A. It helps make prediction

vladimir1956 [14]

The answer is C, it’s a pattern that we can see and use on all right triangles.

3 0

2 years ago