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alexgriva [62]
3 years ago
12

Which of the following is the main evidence of life in the early universe?

Chemistry
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

A)photosynthetic bacteria

alukav5142 [94]3 years ago
3 0
<span>A)photosynthetic bacteria</span>
You might be interested in
If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

<u>Step 3: </u>Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]  

[Ca2+]= 0.050 M   [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

<u> Step 7:</u> Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0
3 years ago
A gas occupies the volume of 215ml at 15C and 86.4kPa?
kherson [118]

Answer:

About 0.1738 liters

Explanation:

Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

0.8527\cdot 0.215=n \cdot 0.0821 \cdot 288\\n\approx 7.754 \cdot 10^{-3}

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

1V=7.754 \cdot 10^{-3} \cdot 0.0821 \cdot 273\\V\approx 0.1738L

Hope this helps!

3 0
3 years ago
7.5 moles of nitrogen gas (N2) is formed in the following reaction. How many grams
AfilCa [17]

Answer:

Mass = 255 g

Explanation:

Given data:

Number of moles of nitrogen = 7.5 mol

Mass of ammonia formed = ?

Solution:

Chemical equation:

3H₂ + N₂      →    2NH₃

Now we will compare the moles of nitrogen and ammonia.

             N₂         :        NH₃

               1          :         2

              7.5       :       2/1×7.5 = 15

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 15 mol × 17 g/mol

Mass = 255 g

6 0
2 years ago
HRISTINA HERRERA: Attempt 1
Lunna [17]

Answer:

7.5 g of hydrogen gas reacts with 50.0 g oxygen gas to form 57.5 g of water.

Explanation:

Here we have the check if the mass of the reactants is equal to the mass of the products.

Reactants

7.5+50=57.5\ \text{g}

Products

57.5\ \text{g}

The data is consistent with the law of conservation of matter.

Reactants

50+243=293\ \text{g}

Products

206+97=303\ \text{g}

The data is not consistent with the law of conservation of matter.

Reactant

17.7+34.7=52.4\ \text{g}

Products

62.4\ \text{g}

The data is not consistent with the law of conservation of matter.

Only the first data is consistent with the law of conservation of matter.

8 0
2 years ago
How many lone pairs of electrons in lewis structure of CH3 OH
Brums [2.3K]
There would be 2 which would be on the oxygen
4 0
2 years ago
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