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Anettt [7]
3 years ago
8

When roasting marshmallows, is the marshmallow the system or the surrounding

Chemistry
2 answers:
Karolina [17]3 years ago
7 0
The marshmallow would be the system
Olenka [21]3 years ago
7 0
The marshmallow should be the system
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A frying pan is connected to a 110-volt circuit. If the resistance of the frying pan is 10 ohms, how many amperes does the fryin
iogann1982 [59]

Answer:

<u>11 A</u>

Explanation:

⇒ V = IR

⇒ 110 V = I * 10 Ω

⇒ I = 110/10

⇒ I = <u>11 A</u>

6 0
3 years ago
In any chemical compound , the elements are always combined in the same proportion by
uranmaximum [27]
The Law of Definite Proportions/Proust's Law
6 0
3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
Read the statement.
lina2011 [118]

Answer:

60 g/L is the final concentration of NaI solution .

Explanation:

Molarity of NaI solution before evaporation =M_1= 0.2 M

Volume of NaI solution before evaporation =V_1= 2.0 L

Molarity of NaI solution after evaporation =M_2= ?

Volume of NaI solution after evaporation =V_2= 1.0 L

M_1V_1=M_2V_2 ( dilution)

M_2=\frac{M_1V_1}{V_2}=\frac{0.2 M\times 2.0 L}{1.0 L}=0.4 M

Molar mass of NaI = 150 g/mol

Concentration of NaI after evaporation :

0.4 M × 150 g/mol = 60 g/L

60 g/L is the final concentration of NaI solution .

4 0
3 years ago
The density of a silver is always ____
Murljashka [212]
The density of silver is 10.49 g/cm^3
5 0
3 years ago
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