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2 years ago

6th Grade Science question on Mass, Volume, and Density!! I've tried and tried and I can't figure it out!

Chemistry

1 answer:

Sonbull [250]2 years ago

8 0

If you want to find density, do mass divided by the volume and there’s your answer!

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_________ energy is to the fission and fusion within an atom as ________ energy is to the fission of atomic compounds.

xeze [42]

I would say C bc I don’t rlly think it’s A or B

7 0

3 years ago

Read 2 more answers

A 2.0 L container of nitrogen gas had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 1.0 atm

beks73 [17]

Answer:

6.4 L

Explanation:

When all other variables are held constant, you can use Boyle's Law to find the missing volume:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.

P₁ = 3.2 atm P₂ = 1.0 atm

V₁ = 2.0 L V₂ = ? L

P₁V₁ = P₂V₂ <----- Boyle's Law

(3.2 atm)(2.0 L) = (1.0 atm)V₂ <----- Insert values

6.4 = (1.0 atm)V₂ <----- Simplify left side

6.4 = V₂ <----- Divide both sides by 1.0

6 0

2 years ago

A rectangle has a volume of 395cm3. It has a mass of 147g. What is its density? *

LekaFEV [45]

The density of a rectangle : ρ = 0.372 g/cm³

<h3>Further explanation</h3>

Given

The volume of rectangle : 395 cm³

Mass : 147 g

Required

The density

Solution

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

Input the value :

ρ = 147 g : 395 cm³

ρ = 0.372 g/cm³

6 0

3 years ago

Can matter be created or<br> destroyed?<br><br> Yes or No

anzhelika [568]

Answer:

Explanation:

The same amount of matter is present before and after chemical and physical changes. Matter cannot be created or destroyed

5 0

2 years ago

Read 2 more answers

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

3 years ago

6th Grade Science question on Mass, Volume, and Density!! I've tried and tried and I can't figure it out!​

6th Grade Science question on Mass, Volume, and Density!! I've tried and tried and I can't figure it out!