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Soloha48 [4]
3 years ago
6

How many milliliters of a 1.25 molar HCL solution would be needed to react completely with 60 g of calcium metal

Chemistry
1 answer:
Svet_ta [14]3 years ago
3 0
Ca + 2HCl = CaCl₂ + H₂

m(Ca)=60 g
c=1.25 mol/l
M(Ca)=40g/mol
v-?

m(Ca)/M(Ca)=m(HCl)/[2M(HCl)]=n(HCl)/2

n(HCl)=2m(Ca)/M(Ca)

n(HCl)=cv

cv=2m(Ca)/M(Ca)

v=2m(Ca)/{cM(Ca)}

v=2·60g/mol/{1.25mol/l·40g/mol}=2.4 l = 2400 ml

2400 milliliters of a 1.25 molar HCl solution would be needed

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JulsSmile [24]

Lava that formed in these columns of Iceland is basaltic lava.

<h3>What is molten basaltic lava?</h3>

The Icelandic type causes the effusions of molten basaltic lava that flow from long, parallel fissures. Such outpourings produce lava plateaus. mountain type of mountains are formed due to such type of eruption of lava. The mountain was formed when the eruption occur under the ice sheet. Basaltic lava is another name of mafic lava. Mafic lava is molten rock that contain iron, magnesium and lower amount of silica. When mafic lava cools on the earth's surface, it produces basalt.

So we can conclude that Lava that formed in these columns of Iceland is basaltic lava.

Learn more about lava here: brainly.com/question/4871898

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6 0
2 years ago
How do you think the human population has changed over time? What makes you say that?
kozerog [31]

Answer:The world population increased from 1 billion in 1800 to 7.7 billion today. The world population growth rate declined from 2.2% per year 50 years ago to 1.05% per year. Other relevant research: World population growth – This article is focusing on the history of population growth up to the present.

Explanation:

3 0
3 years ago
How many molecules are in 4.92 moles of water?
bezimeni [28]

Answer:

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4 0
3 years ago
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
Alexeev081 [22]

Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

Thus, solving for the moles of the acid, we obtain:

n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

Regards.

7 0
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