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Bas_tet [7]
3 years ago
7

How would the freezing point of seawater compare with that of pure water? a)higher b) lower c) the same d) ten times higher

Chemistry
2 answers:
melamori03 [73]3 years ago
4 0
The answer would be B) Lower, because pure water freezes at 32 degrees Fahrenheit whereas salt water freezes at 28.4 degrees Fahrenheit. Hopefully this helps!
olga_2 [115]3 years ago
3 0

Answer:

Freezing point of seawater is lower as compared to pure water

Explanation:

Seawater contains salts like MgCl_{2}, NaCl etc.

According to colligative properties of molecules, presence of solute in a pure solvent lowers freezing point of pure solvent obeying the following equation-

                                        \Delta T_{f}=k_{f}.m

where \Delta T_{f} is lowering of freezing point, k_{f} if cryogenoscopic constant of solvent and m is molality of solution.

In seawater, water is the solvent and MgCl_{2}, NaCl etc are solutes.

Hece freezing point of seawater is lower than pure water

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Immune System is the answer.
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3 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
How does the law of conservation of mass apply to this reaction: Mg + HCl > H2 + MgCl2 ?​
Gelneren [198K]

Answer:

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Chemical equation:

Mg + HCl → H₂ + MgCl₂

24 g + 36.5 g = 2 g+ 95 g

60.5 g = 97 g

The reaction does not hold the law of conservation of mass, because it is not balanced.

Balanced chemical equation:

Mg + 2HCl → H₂ + MgCl₂

24 g + 73 g = 2 g+ 95 g

97 g = 97 g

this equation completely follow the law of conservation of mass.

7 0
3 years ago
If a bar magnet's neutral region is broken into two, what will most likely occur? A. Neither segment will have a north or south
grigory [225]
The answer is D I JUST TOOKM THE TEST
8 0
3 years ago
Read 2 more answers
How many moles are in 297g of nh3?<br><br> Please show work, will give brainliest.
rusak2 [61]

Answer:

17.5moles

Explanation:

The number of moles in a substance can be calculated by using the formula;

Number of moles (n) = mass (m) ÷ molar mass (MM)

According to this question, mass of ammonia (NH3) = 297g

Molar Mass of NH3 = 14 + 1(3)

= 17g/mol

n = 297/17

n = 17.47

Number of moles of NH3 = 17.5moles

3 0
3 years ago
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