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2 years ago

What would happen if an atom of fluorine (F) came into contact with an atom of neon (Ne)?

Chemistry

1 answer:

juin [17]2 years ago

7 0

A. nothing neon is a noble gas, meaning that it dosnt react with anything

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How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid? (4 points) You mus

stiks02 [169]

Balanced Equation: H2SO4 + 2NaOH --> 2H2O + Na2SO4

Moles= Mass/RMM
= 65.5/40
= 1.6375

Mole Ratio = 2:2
= 1.6375

Mass (H2O) = 1.64 x 18
= 29.5 g

5 0

2 years ago

Is density defined as the volume in a given mass orthe mass in a given volume?

wariber [46]

Answer:

mass in a given volume

Explanation:

Density is the mass in a given volume or mass per unit volume of a substance.

Mass is defined by the amount of matter present in a substance.

Volume is the space occupied by a body.

Density is the mass divided by the volume

Mathematically;

Density = $\frac{mass}{volume}$

3 0

2 years ago

Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

2 years ago

How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel

Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

6 0

2 years ago

25.00 gal/min to L/s, Express your answer using four significant figures.

PtichkaEL [24]

Hope this helps you..

6 0

2 years ago