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2 years ago

a one gram sample of hematite up more space than a one gram sample of malachite. what mineral characteristic does this refer to

1 answer:

7 0

I think the mineral characteristics that the one gram sample of hematite taking up more space than a one gram sample of malachite is DENSITY.

The density of hematite is 5.26 g/cm³
The density of malachite is 3.6 to 4 g/cm³

Other physical characteristics of minerals are Color, Streak, Luster, Hardness, Cleavage, Fracture, Tenacity, and Crystal Habit.

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A balloon is floating around outside your window. The temperature outside is 21 ∘C , and the air pressure is 0.700 atm . Your ne

Andreyy89

Answer:

$V=162L$

Explanation:

Hello,

In this case, we can compute the required volume by using the ideal gas equation as shown below:

$PV=nRT$

Thus, solving for the volume and considering absolute temperature (in Kelvins), we obtain:

$V=\frac{nRT}{P}= \frac{4.70mol*0.082\frac{atm*L}{mol*K}*(21+273)K}{0.700atm} \\\\V=162L$

Best regards.

4 0

3 years ago

0.000431 L Express your answer as an integer.

Molodets [167]

A whole number, not a fraction, that can be negative, positive or zero are integers. They cannot have decimal places.

Now, converting 0.000431 L to decimal an integer as:

$0.000431 L = 431\times 10^{-6} L$

Since, $1 L = 10^{6} \mu L$

So, $431\times 10^{-6} L\times \frac{10^{6 \mu L}}{1 L} = 431 \mu L$ .

Hence, the integer value for 0.000431 L is $431 \mu L[$ .

3 0

2 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

2 years ago

Plz answer plzplzpzlpl

astraxan [27]

Whats the question tho?

8 0

2 years ago

What is the name of the compound N2(g)

zavuch27 [327]

N2 stands for nitrogen.
this compound also includes the exact mass.
If you have any other questions please contact me here on Brainly.com and i will be happy to help.
please note- I didn't quite understand the question fully.
-Diane

7 0

3 years ago