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3 years ago

Please help. Idk what to do. Please

Chemistry

1 answer:

Inga [223]3 years ago

5 0

Answer:

7. orbit

8. A comet

9.gas giant

10.moons

11.gravity

12.meteoroids

13.craters

14.super sized

Explanation: idk if 14 is right but yea your welcome

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Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km

miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if 1 g is equal to 100 cg

then 3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if 1 km is equal to 1000 m

then Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

Learn more about:

converting units of measurement

brainly.com/question/11300981

#learnwithBrainly

5 0

4 years ago

An aqueous solution of potassium sulfate (K2SO4) has a freezing point of -2.24

iragen [17]

An aqueous solution of potassium sulfate exhibits colligative properties. Colligative properties are properties that depends on the concentration of a substance in a solution. These properties are freezing point depression, vapor pressure lowering, osmotic pressure and boiling point elevation. For this problem we use the concept of freezing point depression since we are given the freezing point of the solution. Freezing point depression is as:

ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
m = 0.4014

Thus, the molality of the solution is 0.4014.

7 0

4 years ago

A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chl

solong [7]

The reaction between $m_{2}SO_{4}$ with calcium chloride can be shown as- $m_{2}SO_{4}$ + $CaCl_{2}$ → $CaSO_{4}$ ↓+2mCl. The molecular weight of $CaSO_{4}$ is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of $m_{2}SO_{4}$ = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles $m_{2}SO_{4}$ to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989× $10^{-3}$ moles of calcium sulfate.

Thus, 9.989× $10^{-3}$ moles of $m_{2}SO_{4}$ reacts in this reaction.

Let assume the atomic mass of m is x thus the molecular weight of $m_{2}SO_{4}$ is 2x+96.

So we may write 9.989× $10^{-3}$ × (2x+96) =1.42

Or, 2x + 96 = 142.146

Or, 2x = 46.146

Or, x = 23.073

Thus the atomic mass of m is 23.073. The atom (m) is sodium (Na).

8 0

3 years ago

1. Atoms are

natita [175]

Answer:

a.the smallest building blocks of matter

Explanation:

because atoms are the smallest building blocks of matter.

3 0

3 years ago

Read 2 more answers

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKa NH₃/NH₄⁺

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

Moles NH₃

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

H-H equation:

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

3 0

4 years ago