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3 years ago

Which sequence describes the result of increasing the number of gas particles in a container?

Chemistry

1 answer:

geniusboy [140]3 years ago

6 0

Answer:

Explanation:

Adding more gas particles to a set volume will increase the number of collisions, thus increasing collision force and pressure.

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Which one of the following ionic solids would have the largest lattice energy?! A) CaCl2 B) CaBr2 C) Csi D) NaCl E) NaF

Kisachek [45]

The answer would be d

4 0

3 years ago

Which of the following is not a reason why actual yield is less than theoretical yield?

Gnom [1K]

Your answer is B, conservation of mass

Recall that percent yield is given by: %Yeild = actual yeild/theoretical yeild x100

During experiments, there are errors made:

• uncertainty in measurements

• losses of reactants and products

• impurity in reactants

• losses during separation (e.g. filtration or purification)

• Some side reactions might also happen.

Among the given options, only conservation of mass does not contribute to a lower actual yield compared to the theoretical yield.

5 0

3 years ago

An object was measured by a worker as

Alexeev081 [22]

Answer:.633

Explanation:

I have know idea but it was right

3 0

3 years ago

Unit: Chemical Quantities

Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

3 0

4 years ago

Calculate the mass percent (m/m) of a solution prepared by dissolving 45.09 g of NaCl in 174.9 g of H2O.

inysia [295]

Answer:

20.3 % NaCl

Explanation:

Given data:

Mass of solute = 45.09 g

Mass of solvent = 174.9 g

Mass percent of solution = ?

Solution:

Mass of solution = 45.09 g + 174.9 g

Mass of solution = 220 g

The solute in 220 g is 45.09 g

220 g = 2.22 × 45.09

In 100 g solution amount of solute:

45.09 g/2.22 = 20.3 g

Thus m/m% = 20.3 % NaCl

5 0

3 years ago