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3 years ago

Need help with this.

Chemistry

2 answers:

Kay [80]3 years ago

7 0

Answer:

Explanation:

H3PO4.

MA_775_DIABLO [31]3 years ago

6 0

Answer:

I'm pretty sure it's H3PO4

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Consider the following reaction. Upper C upper O subscript 2 (g) plus upper H subscript 2 (g) right arrow Upper C upper O (g) pl

alex41 [277]

Answer:

Its (d) hydrogen

Explanation:

just took the test on edg2020

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3 years ago

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The reaction A + 2B + C -- >D + 2E is first order in reactant A, first order in B, and second order in C. What is the rate la

kvasek [131]

The answer should be D. A rate law needs to be rate equaling the rate constant which is represented as k (make sure you use a lower case k since an upper case K is for equilibrium) times the concentrations of each reactant raised to the power of what ever order it has. (if A was a zero order it would be [A]⁰ and if A was third order it would be [A]³).
Do not get the order the reactants are confused with the coefficients in the chemical equation. (just because the reaction has 2B does not mean the rate law will have [B]². As shown in this example since it is first order therefore being [B] in the rate law)

I hope this helps. Let me know if anything is unclear in the comments.

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3 years ago

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A 0.055 mol sample of formaldehyde vapor, CH2O, was placed in a heated 500 mL vessel and some of it decomposed. The reaction is

Elis [28]

Answer:

Kc for this reaction is 0.06825

Explanation:

Step 1: Data given

Number of moles formaldehyde CH2O = 0.055 moles

Volume = 500 mL = 0.500 L

At equilibrium, the CH2O(g) concentration = 0.051 mol

Step 2: The balanced equation

CH2O <=> H2 + CO

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CH2O] = 0.055 moles . 0.500 L

[CH2O] = 0.11 M

[H2] = 0M

[CO] = 0M

Step 4: The concentration at the equilibrium

[CH2O] = 0.11 - X M = 0.051 M

[H2] = XM

[CO] = XM

[CH2O] = 0.11 - X M = 0.051 M

X = 0.11 - 0.051 = 0.059

[H2] = XM = 0.059 M

[CO] = XM = 0.059 M

Step 5: Calculate Kc

Kc = [H2][CO]/[CHO]

Kc = (0.059 * 0.059) / 0.051

Kc = 0.06825

Kc for this reaction is 0.06825

7 0

3 years ago

Nitrogen is an atom or molecules

Pavel [41]

Answer:

Nitrogen will called as atom or molecule or ion too in the state which it exist means in which form it is present .

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2 years ago

112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.

dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

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

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

$\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol$

For the given chemical reaction:

$2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)$

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = $\frac{12}{2}\times 0.778=4.668 mol$ of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

$\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol$

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

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2 years ago