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3 years ago

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Acetylene, C2H2, burns according to the following reaction: C2H2 5O2 --> 4CO2 2H2O. Suppose 1.20 g of C2H2 is mixed with 3.50

g of O2 in a closed, steel container, and the mixture is ignited. What substances will be found in the mixture left when the burning is complete

1 answer:

OLEGan [10]3 years ago

8 0

WWWWWWAAAAAAPPPPPP!!!!!

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2. Write the chemical equations for the neutralization reactions that occurred when HCL and NaOH were added to the buffer soluti

lutik1710 [3]

Answer:

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Explanation:

A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.

A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equations for the neutralizations that occurred upon addition of HCl or NaOH are;

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

5 0

3 years ago

what would the molarity of a solution be if you took 10 mL of a 13M stock solution and made a 300 mL solution?

nignag [31]

Answer:

0.43M

Explanation:

8 0

3 years ago

When sodium bicarbonate and acetic acid are combined in a beaker and allowed to

777dan777 [17]

Answer: C excess reactant

Reason: It is not catalyst because catalyst are not consumed. It is not product because it was one of the reactants. It is not limiting because there was excess left over.

8 0

3 years ago

Read 2 more answers

Help please you don’t need to show work

aleksandrvk [35]

Answer:

24.07

Explanation:

6 0

3 years ago

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as

erica [24]

<u>Answer:</u> The amount of $P_4O_{10}$ formed is 469.8 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol$

The given chemical reaction follows:

$4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)$

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of $P_4O_{10}$

So, 6.62 moles of phosphine will produce = $\frac{1}{4}\times 6.62=1.655mol$ of $P_4O_{10}$

Now, calculating the mass of $P_4O_{10}$ by using equation 1:

Molar mass of $P_4O_{10}$ = 283.9 g/mol

Moles of $P_4O_{10}$ = 1.655 moles

Putting values in equation 1, we get:

$1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g$

Hence, the amount of $P_4O_{10}$ formed is 469.8 grams.

8 0

3 years ago