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timama [110]
1 year ago
5

When choosing a respirator for your job, you must conduct a _____ test.

Chemistry
1 answer:
Lina20 [59]1 year ago
4 0
(B) breathing (test)
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A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevatio
Goryan [66]

Answer:

  MCl₂

Explanation:

The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.

  \Delta T=imK_b\qquad\text{where i=ions/mole, m=molality, $K_b\approx 0.512$}\\\\376.81-373.15=(x+1)\dfrac{\text{moles}}{\text{kg solvent}}(0.512)\\\\\dfrac{3.66}{0.512}=(x+1)\dfrac{\dfrac{30.2}{56+35.45x}}{0.1}=\dfrac{302(x+1)}{56+35.45x}\\\\\dfrac{3.66}{0.512\cdot 302}(56+35.45x)=x+1\\\\\dfrac{3.66\cdot 56}{0.512\cdot 302}-1=x\left(1-\dfrac{3.66\cdot 35.45}{0.512\cdot 302}\right)\\\\x=\dfrac{50.336}{24.877}\approx 2.023

Then the formula for the salt is MCl₂.

6 0
1 year ago
Dr. Franck cuts a bar of pure gold into smaller and smaller pieces. Will this action change the element that makes up the bar? E
Anvisha [2.4K]

Answer:

i believe this is a chemical or physical question? well your answer to that is no the element does not change because the gold is still gold it is still physical because you have just cut it into piece it is still gold

Explanation:

lmk if it was helpful :/

7 0
1 year ago
How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?
maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is PbCrO_{4} and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Putting the values,

n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0
1 year ago
Imagine if during the cathode ray experiment, the size of the particles of the ray was the same as the size of the atom forming
Yakvenalex [24]

Answer:

This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.

Explanation:

Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.

Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.

If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.

7 0
2 years ago
Read 2 more answers
The combustion of propane is represented by the following chemical equation. C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l) The standard enthalp
wariber [46]

Answer:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

7 0
2 years ago
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