Ask question

Ask question

All categories

DanielleElmas [232]

3 years ago

7

A.1 B.2 C.3 D.4 Please help!

2 answers:

VashaNatasha [74]3 years ago

6 0

I think its 3 precipitation.

Lena [83]3 years ago

6 0

The answer is a i think

You might be interested in

Why does heating a liquid affect viscosity

sveta [45]

Answer:

In general, liquids tend to get “thinner” when their temperature increases. For example, honey and oil tend to flow better at higher temperatures. Therefore, increasing temperature decreases viscosity. In general, the liquids tend to expand when their temperature increases

Explanation:

7 0

3 years ago

This method of treating water kills microbes and removes other contaminates such as heavy metals.

Ivenika [448]

The answer is (B).
Hope this helps :).

3 0

3 years ago

A certain gas is present in a 15.0 LL cylinder at 2.0 atmatm pressure. If the pressure is increased to 4.0 atmatm the volume of

Helga [31]

Explanation:

According to Boyle's law, pressure of a gas is inversely proportional to its volume at constant temperature and moles.

Mathematically, P = $\frac{k}{V}$

where, k = proportionality constant

Also, formula for initial pressure and volume is as follows.

$P_{i} = \frac{k_{i}}{V_{i}}$

or, $k_{i} = P_{i} \times V_{i}$

= $2 atm \times 15 L$

= 30 atm L

Now, we will calculate the value of $k_{f}$ as follows.

$k_{f} = P_{f} \times V_{f}$

= $4 atm \times 7.5 L$

= 30 atm L

Hence, as $k_{i} = k_{f}$ this means that it signifies that gas obeys boyle's law.

6 0

3 years ago

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of

Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

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

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

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

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the alloy is 2.324 J/g°C

3 0

3 years ago

Can somebody plz help answer all of these questions right it’s very important. If u can that will be amazing thanks!

Blababa [14]

Answer:

1. Mechanical energy

2. Heat/thermal energy

3. Electrical energy

4. Heat energy

5. Electrical energy

Explanation:

Hope this helped!

4 0

3 years ago