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3 years ago

Oi oi oi oi ewin .................

Chemistry

1 answer:

KatRina [158]3 years ago

7 0

Muy bein me gusta hope this makes you smile

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Oxidation number of Mg3N2

Fudgin [204]

Answer:

n=-3

Explanation:

Mg = +2

3(+2) + 2N = 0

N = -3

8 0

4 years ago

Help me with please !!!

egoroff_w [7]

I would think the last one about the ozone layer

6 0

3 years ago

A water sample has a pH of 8.2 and a bicarbonate concentration of 97 mg/L. What is the alkalinity of the sample in moles/L and i

goldfiish [28.3K]

Answer:6.94

Explanation:

Molar mass of CaCO3=40+12+16×3

=40+12+48=100g/mol

Moles=mass of substance/molar mass

=97mg/100g=0.097/100=0.00097moles/L.

PH=-log[CaCo3]=-log(0.00097)=6.94

P.s it's log to base e

3 0

3 years ago

Uranium-234 is unstable and undergoes two subsequent alpha emissions. What is the resulting nuclide from this transformation?

MArishka [77]

Answer:Radium

Explanation:

The nuclear reaction involving two alpha emissions of 234 U is shown in the diagram. This leads to the formation of a 226Ra nucleus.

4 0

3 years ago

A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur

lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g$

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$q=m\times c\times \Delta T$

q = heat absorbed by water

$m$ = mass of water = 40.0 g

$T_{final}$ = final temperature of water = 20.0°C

$T_{initial$ = initial temperature of water = 17.0°C

$c$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$q=40.0\times 4.186\times (20.0-17.0)]$

$q=502J$

Hence, the joules of heat were re-leased by the lead is 502

5 0

3 years ago