Which organisms have greater variation and produce offspring through asexual reproduction?

If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch

Evgesh-ka [11]

Answer:

0.120M is the concentration of the solution

Explanation:

<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>

<em />

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

<em>Molar concentration:</em>

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>

7 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

Allisa [31]

Answer:

The mass of water $m_{w}$ = 39.18 gm

Explanation:

Mass of iron $m_{iron}$ = 32.5 gm

Initial temperature of iron $T_{1}$ = 22.4°c = 295.4 K

Specific heat of iron $C_{iron}$ = 0.448 $\frac{KJ}{kg K}$

Mass of water = $m_{w}$

Specific heat of water $C_{w} = 4.2 \frac{KJ}{kg K}$

Initial temperature of water $T_{2}$ = 336 K

Final temperature after equilibrium $T_{f}$ = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water = Heat gain by iron rod

$m_{w}$ $C_{w}$ ( $T_{2}$ - $T_{f}$ ) = $m_{iron}$ $C_{iron}$ ( $T_{f}$ - $T_{1}$ )

Put all the values in above formula we get

$m_{w}$ × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

$m_{w}$ = 39.18 gm

Therefore the mass of water $m_{w}$ = 39.18 gm

8 0

3 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

In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical

Umnica [9.8K]

Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5

3 0

3 years ago

During oxidation, which of the following is true? A. Electrons are gained, so the oxidation number increases. B. Electrons are g

EastWind [94]

Oxidation is when the overall charge (or oxidation number) increases. The only way to increase an oxidation number is to lose an electron, thereby making the negative charges less. The correct answer is C.

6 0

3 years ago

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