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Neko [114]
3 years ago
7

A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumet

ric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.
Chemistry
1 answer:
frutty [35]3 years ago
6 0

Answer:

0,12 μmol/L of MgF₂

Explanation:

Preparation of solutions is a common work in chemist's life.

In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in  μmol/L

You have 0,00598 μmol but not Liters.

To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:

50,0 mL (1L/1000mL) = 0,05 L of water.

Thus, concentration in  μmol/L is:

0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-

I hope it helps!

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a sample of the hydrate of sodium carbonate has a mass of 8.85 g. it loses 1.28 g when heated. find the molar ratio of this comp
xenn [34]

Answer:

Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.

Explanation:

From the given,

Mass of sodium carbonate Na_{2}CO_{3}.XH_{2}O = 8.85 g

Loss mass H_{2}O = 1.28 g

Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g

7.57 g Na_{2}CO_{3} \times \frac{1mol}{106 g} =\frac{0.0714}{0.0714}=1

1.28g H_{2}O \times \frac{1mol}{18 g} =\frac{0.0711}{0.0714}=1

Therefore, the compound has only one water molecule.

Molecular formula of the compound is Na_{2}CO_{3}.H_{2}O an name of the compound is <u>sodium carbonate mono hydrate.</u>

Hence, the type of the compound is Mono hydrate.

8 0
3 years ago
What is the salinity seawater?
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7 0
3 years ago
Which best describes the effect of J. J. Thomson’s discovery?
vlabodo [156]
The correct answer for the question that is being presented above is this one: "A. The accepted model of the atom was changed.<span>" </span>J J Thomson discovered the electrons and performed experiment using the cathode ray tube
Here are the following choices:
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8 0
3 years ago
What mass of salt would you need to add to 1.00kg of water to achieve a freezing point of -5 degrees C
Serga [27]

Answer:

The type of salt to be added to the water is not known from the question but no worries, I will try to give you the step by step procedure to answer any type of question similar to this.

To answer this question, we should know some facts.

1. the molar freezing point depression constant of water (Kf) = 1.86 K kg/mol

2. the molar mass of the salt if NaCl = 58.5 g/mol ; KCl = 74.5 g/mol

3. since the salt can dissociate if NaCl or KCl into two ions, the Van't Hoff factor ( i )= 2

Note that: the change in freezing point, molarity, deepression constant and van't Hoff factor are related by this formula;

                ΔTf = i Kf m

So lets take NaCl as the salt:

Molar mass = 58.5 g/mol

Van't Hoff factor = 2

1. calculate the number of moles

So we can calculate the molarity of the salt NaCl from the formula;

m = ΔTf / i Kf

m = 5 / 2 * 1.86

m = 5 / 3.72

m = 1.344 mol/kg

2. calculate the number of moles of the salt required

Next is to multiply the molarity by the mass of water. Density of water = 1kg/L

number of moles = 1.344 mol/kg * 1 Kg/L * 1 kg water

number of moles = 1.344 moles.

3. calculate the mass of the salt.

numner of moles = mass / molar mass

mass = number of moles * molar mass

mass = 1.344 * 58.5

mass = 78.624 g of NaCl salt.

You can follow these steps to solve for the type of salt you are given in the question.

6 0
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Tema [17]

Answer:

gravity

Explanation:

Gravity is an example of the force of attraction between two objects that have mass.

3 0
3 years ago
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