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Maurinko [17]
3 years ago
15

PLEASE HELP I BEG YOU I REALY NEED HELP

Chemistry
2 answers:
Korvikt [17]3 years ago
5 0

Answer: 1. Physical change

2. volume

3. true

Explanation:

padilas [110]3 years ago
3 0

Answer: 1

Explanation:

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Lelu [443]

Answer: See attached image

Explanation:

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3 years ago
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For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate sol
diamong [38]

Answer: ksp= 4s³

Explanation:

4 0
3 years ago
Iron has a known density of 7.87 g/cm3. What would be the mass of a 2.5<br> cm3 piece of iron?
Fudgin [204]
If i am not mistaken ,

density = mass / volume

density = 7.87 g/cm3
volume = 2.5 cm3

thus ,
mass = density x volume

7.87 g/cm3 x 2.5 cm3 = 19.675 g
5 0
3 years ago
A tactic used by investigators to identify deceit is to ask the suspect to recount their alibi backwards. True False
MA_775_DIABLO [31]

Answer:

It is True.

Explanation:

Investigators use this tactic to see if the suspect is recalling by memory and if they will change their story after being asked to recall backwards.

8 0
3 years ago
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Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0 C, the pr
Paladinen [302]

The question is incomplete, here is the complete question:

Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.230 atm. What is the likely formula of the compound?

A. XeO

B. XeO_4

C. Xe_2O_2

D. Xe_2O_3

E. Xe_3O_2

<u>Answer:</u> The chemical formula of the compound is XeO_4

<u>Explanation:</u>

To calculate the molecular weight of the compound, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{w}{M}RT

where,

P = Pressure of the gas = 0.230 atm

V = Volume of the gas  = 50.0 mL = 0.050 L     (Conversion factor:  1 L = 1000 mL)

w = Weight of the gas = 0.100 g

M = Molar mass of gas  = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gas = 0^oC=273K

Putting value in above equation, we get:

0.230\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.230\times 0.050}=194.9g/mol\approx 195g/mol

The compound having mass as calculated is XeO_4

Hence, the chemical formula of the compound is XeO_4

5 0
4 years ago
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