I know the photo trash but i need answers for the top middle and last question please :)

Chemistry

1 answer:

Firlakuza [10]2 years ago

6 0

Answer:

Middle: Self heating containers are really expensive but are useful because they are easy to use, portable, and can be recycled, unlike a camp stove which is not as easily moved.

Explanation:

i neeed a better image for the first an last boo<3

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The percent yield of a chemical reaction is valuable to many industries. Low percent yields can result in the loss of large amou

aev [14]

Using accurate measurements, using pure chemicals and performing the reaction under the most ideal conditions is important to get a valuable percent yield.

<h3>How we calculate the percent yield?</h3>

Percent yield of any chemical reaction is define as the ratios of the actual yield to the theoretical yield of the product and multiply by the 100.

To get the high percent yield or actual yield of any reaction, we have to perform the reaction under ideal condition because if we not use the standard condition then we get the low rate of reaction. Reactants should be present in the pure form as impurity make unwanted products and reduce the productivity of main product and accurate amount of reactants also important for the spontaneous reaction.

Hence, options (a), (b) & (c) are correct.

To know more about percent yield, visit the below link:

brainly.com/question/8638404

7 0

2 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}}$