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3 years ago

What type of mixture will not allow light to pass through?

2 answers:

8 0

Anything that won't let light through is opaque. If it lets some light through, but not all, it is translucent. And if it lets all of the light through, it is transparent.

stealth61 [152]3 years ago

3 0

Suspensions. The size of the particles are usually big enough to see, and will separate when left to sit.

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Predict: If the skater starts at a height of 2m on one side of the ramp and starts to move, how high will he get on the other si

OverLord2011 [107]

Answer:

Sliding friction

Explanation:

SLiding friction produces force

7 0

3 years ago

Within a period, the nuclear charge increases as you move from left to right across the

a_sh-v [17]

Answer:

The answer is Ionization energy.

Explanation:

Ionization Energy. The ionization energy tends to increase as one moves from left to right across a given period or up a group in the periodic table.

3 0

3 years ago

Find the mass of 5.82 mol of MgCl2. Round to the nearest tenth.

Oksana_A [137]

Answer:

554.13 grams

Explanation:

8 0

3 years ago

Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h

Gwar [14]

Answer:

0.264 g of $CO_2$ can be formed from 288 mg of $O_2$

Explanation:

The balanced chemical equation is

$2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O$

The conversions are

Mass in mg $O_2$ is converted to mass in g $O_2$

Mass in g $O_2$ is converted to moles $O_2$ by dividing with molar mass

Moles $O_2$ is converted to moles $CO_2$ by using the mole ratio of $O_2:CO_2$ is 9 : 6

Moles $CO_2$ is converted to mass $CO_2$ by multiplying with molar mass $CO_2$

mass in mg $O_2$ > mass in g $O_2$ >moles $O_2$ > moles $CO_2$ > mass $CO_2$

$288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}$

=0.264g (Answer)

7 0

4 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

3 years ago