All categories

3 years ago

Geckos and some other animals can climb smooth surfaces. Which of these explanations of this

Chemistry

2 answers:

Montano1993 [528]3 years ago

8 0

Answer:

c) pillars

Explanation:

Marrrta [24]3 years ago

4 0

Answer:

The gecko's feet have tiny suction cups on them that stick to the surface.

You might be interested in

The release of a gas from a chemical reaction is very quick. A scientist wants to measure what is released, but in small increme

Ganezh [65]

An inhibitor, which slows down the reaction enough to measure the release of gas.

5 0

3 years ago

Read 2 more answers

A helium filled weather balloon has a volume of 806 L at 20.9°C and 753 mmHg. it is released and rises to an altitude of 6.8 km,

OLEGan [10]

<h3>Answer:</h3>

1257.45 L

<h3>Explanation:</h3>

We are given;

Initial volume of Helium gas, V1 as 806 L
Initial temperature of Helium gas,T1 as 20.9°C
Initial pressure of Helium gas, P1 as 753 mmHg
Pressure of Helium at the altitude 6.8 km, P2 as 417 mmHg
Temperature of Helium gas at the altitude 6.8 Km, T2 as -19.1°C

But, K = °C + 273.15

Therefore, T1 = 294.05 K and T2 = 254.05 K

We are required to calculate the new volume of the balloon at 6.8 km.
To determine the new volume we are going to use the combined gas law.
According to the combined gas law, $\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Thus, rearranging the formula;

$V2=\frac{P1V1T2}{P2T1}$

$V2=\frac{(753)(806L)(254.05K)}{(417)(294.05)}$

$V2=1257.45L$

Therefore, the volume of the balloon at an altitude of 6.8 km is 1257.45 L

7 0

3 years ago

Describe use of H₂S an analytical reagent?<br><br>

Gemiola [76]

Answer:

The main use for hydrogen sulfide is in the production of sulfuric acid and elemental sulfur. ... H2S is used to prepare the inorganic sulfides you need to make those products. As a reagent and intermediate, hydrogen sulfide is beneficial because it can prepare other types of reduced sulfur compounds.

7 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

What must happen for heat to be transferred by convection

Marrrta [24]

Answer: The heat transfer by convection requires a solid-fluid interface, a temperature difference between the solid surface and the surrounding fluid and a motion of the fluid. The process of heat transfer by convection would occur when there is a movement of macro-particles of the fluid in space from a region of higher temperature to lower temperature.

Explanation:

6 0

3 years ago