All categories

3 years ago

Pleas answer this, it’s due tomorrow. I will give brainliest.

Chemistry

2 answers:

Vilka [71]3 years ago

8 0

Answer:

just the second

Explanation:

larisa86 [58]3 years ago

6 0

Answer:

both are changes in state, and both conserve mass. Hope it helps and please have a good day or night

You might be interested in

If the amount of Adams if each type on the left and right sides of a reaction difference what must be done to balance it

Ivanshal [37]

Hey there,

Your answer would be

Coefficients are placed in front of the reactants and/or products

Hope this helps,

<h2>- <em>Mr. Helpful</em></h2>

8 0

3 years ago

A chemist has 3.55x1022 molecules of nitrogen monoxide. How

Alina [70]

Answer:

<h2>0.059 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.55 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.05897...$

We have the final answer as

<h3>0.059 moles</h3>

Hope this helps you

6 0

3 years ago

A sample of methane gas has a volume of 355 mL at 25.0°C and 2414 mm Hg. If the temperature is raised to 100.°C and the pressure

Drupady [299]

Answer:

23019

Explanation:

5 0

3 years ago

Read 2 more answers

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

What evidence is there that energy is conserved?

Tems11 [23]

The law of conservation

Hope this helps

~ Jordan ~

5 0

3 years ago