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Chemistry

1 answer:

Alexus [3.1K]3 years ago

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Answer:

In the case of water, an object with a density less than 1 g/cm3 will float. The closer its density is to 1 g/cm3, the more of it will sit below the water level.

Explanation:

Whether an object will float or sink is dependent on its density, and on the density of the liquid it is placed in.

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Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO

sergey [27]

The molarity of a solution is the number of moles of a substance divided by the volume in liters prepared.

$molarity=\frac{n}{V}$ , where n is number of moles and V is the volume in liters.

In order to calculate the mass of solute we need to convert the volume and molarity to moles

$1.421 L solution \times\frac{0.0730 moles}{1 Lsolution}= 1.037 mol NaCO_3$

Now that we have moles we use the relative formula mass of NaCO₃, We have 1 Na atom, 1 C atom and 3 O atoms, thus

$M_r= (1\times 22.99) + (1\times 12.00) + (3\times 16.00)= 82.99g/mol$

$1.037 \times\frac{82.99g}{mol} = 86.1g$

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3 years ago

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0.40 L of HNO3 is titrated to equivalence using 0.18 L of 0.1 M NaOH. What is the concentration of the HNO3 ?

e-lub [12.9K]

This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

<h3>Acid-base titrations:</h3>

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

$M_AV_A=M_BV_B$

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.