Which statement is true about the following isotopes: Sulfur-34, Sulfur-33 and Sulfur-32? (Choose all that apply)

How is kinetic energy of particles and temperature related

Rufina [12.5K]

Answer:

The higher the temperature, the faster the particles move, the lower the temperature, the slower.

4 0

3 years ago

Find the area of a vegetable garden that is 4.83m long and 3.4m wide. Give your answer to the correct significant figure. (use t

sergejj [24]

Answer:

The area of vegetable garden is 16.422 m².

Explanation:

Given data:

Length = 4.83 m

Width = 3.4 m

Area = ?

Solution:

Formula:

Area = length × width

Area is measured in square unit.

By putting the values in formula:

Area = length × width

Area = 4.83 m × 3.4 m

Area = 16.422 m²

The area of vegetable garden is 16.422 m².

5 0

3 years ago

Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at

WITCHER [35]

<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL

<u>Explanation:</u>

Specific gravity is given by the formula:

$\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}$

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

$0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL$

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

$\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL$

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0

3 years ago

Which of the following is an irreversible physical change?

Yuliya22 [10]

Answer:

b) sharpening a pencil

Explanation:

If you melt lead, boil water, or dissolve sugar in water, you can return all of them back to their original state. If you sharpen a pencil, you can't reattach the shavings as they were originally.

3 0

3 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

3 years ago