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sleet_krkn [62]
2 years ago
15

why does an aluminum horseshoe bend but not break when a blacksmith pounds it into shape with a hammer?

Chemistry
1 answer:
riadik2000 [5.3K]2 years ago
8 0

Answer:

it's because aluminium is a ductile material

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Using your own words whats the difference between weather and climate ​
aksik [14]

Answer:

Weather is at a specific place and time. Climate is over a period of time.

Explanation:

That is the difference

8 0
3 years ago
15.3 g of nano3 were dissolved in 100g of water in a calorimeter. The temperature of the water drops from 25.00°c to 21.56°c. Ca
Arturiano [62]

Answer:

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 100.0 g).

c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

<em>∵ Q = m.c.ΔT</em>

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

<em>∵ ΔH = Q/n</em>

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

3 0
3 years ago
PLEASE HELP, CHEMISTRY!!! 20 POINTS!!!
mart [117]

the answer is in the picture

6 0
3 years ago
Read 2 more answers
Hydrochloride acid +_________ zinc chloride +hydrogen​
Klio2033 [76]

Answer:

Hydrochloride acid + Zinc = Zinc Chloride + Hydrogen

Explanation:

When Hydrochloride acid and Zinc react, it results in the formation of Zinc chloride and hydrogen.

<em>Hope I helped</em>

3 0
2 years ago
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

2*n_{acid}=n_{base}

And in terms of concentrations and volumes:

2*M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the volume of acid:

V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL

Best regards.

5 0
3 years ago
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