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4 years ago

How many neutrons dose O^-2 have

1 answer:

8 0

Oxygen is the 8th element in the periodic table. This means that oxygen has 8 protons<span> and 8 electrons. In order to get the number of neutrons you take the atomic weight in this case 15.9999~16 and you subtract it by the number of protons (16-8) (o_O)

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Bromthymol blue is a ph indicator that turns yellow in the presence of carbon dioxide.

rjkz [21]

A because carbon dioxide is a gas with molecules

6 0

3 years ago

A block of material has a volume of 50 cubic centimeters, and a mass of 1,000 grams; what is that object's density?

zhannawk [14.2K]

Answer:

Density = $20\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 1000 g

Volume of the block, V = 50 cm³

We need to find the density of the object. Density is equal to mass per unit volume.

d = m/V

$d=\dfrac{1000\ g}{50\ cm^3}\\\\=20\ g/cm^3$

So, the density of the object is $20\ g/cm^3$ .

8 0

3 years ago

What mass of hydrogen is produced when 192 g of magnesium is reacted with hydrochloric acid?

mariarad [96]

Answer:

Mg+2HCL-Magnesium Chloride +Hygrogen

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Explanation:

N.O of moles=Mass\Molar Mass

192÷24=8

1:1

8×2=16

3 0

2 years ago

Read 2 more answers

An element has two naturally-occurring isotopes. The mass numbers of these isotopes are 115.00 u and 117.00 u, with natural abun

denis23 [38]

$(115.00)(0.25)+(117.00)(0.75)=\boxed{116.50 \text{ u}}$

3 0

2 years ago

A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sampl

Rzqust [24]

Answer:

0.193 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 0.250 L

Initial temperature (T1) = 27°C

Initial pressure (P1) = 0.850 atm

Final volume (V2) =?

Final temperature (T2) = 0°C

Final pressure (P2) = 1.00 atm

Step 2:

Conversion of celsius temperature to Kelvin temperature.

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 27°C = 27°C + 273 = 300K

Final temperature (T2) = 0°C = 0°C + 273 = 273K

Step 3:

Determination of the new volume of the sample of ammonia gas.

The new volume can be obtain by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

0.850x0.250/300 = 1xV2/273

Cross multiply to express in linear form

300 x V2 = 0.850x0.250x273

Divide both side by 300

V2 = (0.850x0.250x273) /300

V2 = 0.193 L

Therefore, the volume of the sample at 0 °C and 1.00 atm is 0.193 L

6 0

3 years ago