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lyudmila [28]
3 years ago
15

Is it a spontaneous or non-spontaneous change?

Chemistry
1 answer:
joja [24]3 years ago
7 0
Spontaneous is no spontaneous
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What type of decay involves a change in mass?
kaheart [24]
Beta decay is the answer
6 0
2 years ago
Please i need the answer for 18,19,20<br><br><br>Answer and I will give you brainiliest <br>​
dezoksy [38]
Number 18 is correct, number 19 is the particles speed up and the substance expands, and number 20 is decrease in volume
8 0
3 years ago
PLEASE HELP! WILL GIVE BRAINLIEST:
Ad libitum [116K]

2.45 °C

From the Ideal gas law (Combined gas law)

PV/T = P'V'/T' .....eq 1

Where:

P - initial pressure

V - initial volume

T - initial temperature

P' - final pressure

V' - final volume

T' - final temperature.

To proceed we have to make T the subject from eq.1

Which is, T = P'V'T/PV.......eq.2

We have been provided with;

Standard temperature and pressure (STP)

P = 760 mm Hg (SP - Standard Pressure)

T = 273.15 K (ST - Standard Temperature)

V = 62.65 L

P' = 612.0 mm Hg

V' = 78.31 L

T' = ? (what we require)

Therefore, we substitute the values into eq.2

T' =

.

T' = 275.60 K

T = (275.60 - 273.15) ......To °C

T = 2.45 °C

>>>>> Answer

Have a nice studies.

5 0
2 years ago
Correct the name to match the formula of the following compounds: (a) calcium(ll) dichloride. CaClZ; (b) copper(|l) oxide. Cu20;
tangare [24]

Answer:

Explanation:

Incorrect name                                  correct name

a) calcium dichloride                         calcium chloride

b) copper(II) oxide                               copper(I) oxide

c) stannous tetrafluoride                      stannic fluoride

d) hydrogen chloride acid                     hydrochloric acid

5 0
3 years ago
a compound containing only carbon and hydrogen produces 1.80 gCO2 and 0.738 gH2O. Find the empirical formula of the compound. Ex
r-ruslan [8.4K]

Answer:

The empirical formula is CH2

Explanation:

Step 1: Data given

Mass of CO2 produced = 1.80 grams

Mass of H2O produced = 0.738 grams

Step 2: The balanced equation

CxHy + O2 → CO2 + H2O

Step 3: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 1.80 grams / 44.01 g/mol

Moles CO2 = 0.0409 moles

Step 4: Calculate moles C

In 1 mol CO2 we have 1 mol CO2

In 0.0409 moles CO2 we have 0.0409 mol C

Step 5: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 0.738 grams / 18.02 g/mol

Moles H2O = 0.04095  moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.04095 moles H2O we have 2*0.04095 = 0.0819 moles H

Step 7: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.0409 / 0.0409 = 1

H: 0.0819 moles / 0.0409 =  2

The empirical formula is CH2

4 0
3 years ago
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