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3 years ago

What grey rock that has crinoids and brachiopods reacts to hydrochloric acid?

2 answers:

6 0

Hey there Barney456,

What grey rock that has Crinoids and Brachiopods reacts to Hydrochloric acid?

Answer:

Limestone. It is found underwater. When marine organisms die their skeletons compact and form the limestone

Hope this helps :D

<em>~Danielle♥</em>

Anon25 [30]3 years ago

5 0

The rock is limestone, I hope this helps!!

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When an acid or vase is added to to water,

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yes

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2 years ago

Determine whether or not the equation below is balanced. If it isn’t balanced, write the balanced form. Also, identify the react

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Answer:

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3 years ago

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At high pressures and moderate temperatures nitric oxide gas disproportionates rapidly according to the reaction 3NO(g) ⇌ N2O(g)

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Answer:

3.8 x 10⁵

Explanation:

For the equilibrium : 3NO(g) ⇌ N2O(g) + NO2(g), the equilibrium constant in the terms of the concentrations of the gases in mol/L is

Kc = (NO) (N2O)/ (NO) ³ where (NO), (N2O) , (NO2) are the concentrations of the gases in mol/L . So

K= (x mol/ 1 L) (x mol/1L) / (7.5 x 10⁻⁶ mol /1 L) ³

x = mol of NO and NO2 at equilibrium

we have that

K = x²/ 7.5 x 10⁻⁶ = 1.9 x 10¹⁶

x = √ (7.5 x 10⁻⁶ x 1.9 x 10¹⁶) = 3.8 x 10⁵

∴ (N2O) = 3.8 x 10⁵

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3 years ago

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3 years ago

In a gas grill, 29 lbs propane C3H8 are

dimulka [17.4K]

Answer : The mass of combustion products formed are 134 lbs.

Explanation :

The balanced chemical reaction will be:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Given :

Mass of $C_3H_8$ = 29 lbs = 13154.2 g

conversion used : 1 lbs = 453.592 g

Molar mass of $C_3H_8$ = 44 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles$

Now we have to calculate the moles of $CO_2$ and $H_2O$ .

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_3H_8$ react to give 3 moles of $CO_2$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 3=896.7$ moles of $CO_2$

and,

As, 1 mole of $C_3H_8$ react to give 4 moles of $H_2O$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 4=1195.6$ moles of $H_2O$

Now we have to calculate the mass of $CO_2$ and $H_2O$ .

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2$

$\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs$

and,

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O$

$\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs$

The total mass of products = Mass of $CO_2$ + Mass of $H_2O$

The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs

Therefore, the mass of combustion products formed are 134 lbs.

6 0

3 years ago