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3 years ago

When was the death of all dinosuars?

1 answer:

4 0

The Cretaceous-Tertiary extinction event , the name given to the die off dinosaurs, they all died off about 65.5 million years ago . They died by climate and geological changes .

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How many moles of H2O2 are needed to react with 1.07 moles of N2H4?

I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄ + 2H₂O₂ → N₂ + 4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

N₂H₄ : H₂O₂

1 : 2

1.07 : 2×1.07 = 2.14 mol

6 0

3 years ago

Write out the word equations for these reactions. ) Hydrogen reacts with oxygen to form water.

ale4655 [162]

Answer:

H2O

Explanation:

PLS MARK ME TO THE BRAINLIST

4 0

3 years ago

A new substance is formed when the?

quester [9]

Answer:

A chemical reaction happens when substances break apart or combine to form one or more new substances.

Explanation:

hope its right.

4 0

3 years ago

What is the chemical formula for the ternary compound composed of co3+ and so42- ions? coso4 co3(so4)2 co2so4 co2(so4)3 none of

Andreas93 [3]

If you are provided with Cation and an Anion with different oxidation states, then there ratio in the formula unit is adjusted as such that the oxidation number of one ion is set the coefficient of other ion and vice versa,

Example:
Let suppose you are provided with A⁺² and B⁻¹, so multiply A by 1 and B by 2 as follow,
A(B)₂

In statement we are given with Co⁺³ and SO₄⁻², so multiply Co⁺³ by 2 and SO₄⁻² by 3, hence,

Co₂(SO₄)₃

Result:
Co₂(SO₄)₃ is the correct answer.

4 0

3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Mama L [17]

Answer: The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

Explanation:

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

where,

$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

Hence, the number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

5 0

3 years ago