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3 years ago

Scientist discover that stars in the universe are mostly composed of which elements?

Chemistry

1 answer:

konstantin123 [22]3 years ago

3 0

Scientists discovered that starts in the universe are comprised mostly of hydrogen and helium, I believe.

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What is the definition of a chemical bond?

Gnom [1K]

Answer:

Explanation:

bonding is a process of two different atoms sharing electrons for stability and these electrons are attracted by one atom losing it's electrons to another

7 0

2 years ago

N2H4+H2O2 ——> N2+H2O coefficient

const2013 [10]

Answer:

N2H4 + 2H2O2 ---->N2 + 4H2O

Explanation:

N=2 N=2

H=6 ->8 H=2 ->8

O=2 -> 4 O=1 -> 4

Add coefficients to hydrogen peroxide on the left and water on the right, so that there is an equal number of hydrogens and oxygens.

3 0

2 years ago

The first strings of holiday lights were manufactured as series circuits. Which explains why this method is no longer used? A. I

user100 [1]

B because once the circuit is burned the lights will all go of

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2 years ago

3.8 liters of sulfur vapor, S8(g), at 921.4°C and 5.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measur

lilavasa [31]

Answer:

116.5 g of SO₂ are formed

Explanation:

The reaction is:

S₈(g) + 8O₂(g) → 8SO₂ (g)

Let's identify the moles of sulfur vapor, by the Ideal Gases Law

We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K

5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K

(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈

Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide

Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 = 1.82 moles

We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g

3 0

2 years ago

Read 2 more answers

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0

3 years ago