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3 years ago

11

a substance which alters the rate of a chemical reaction but is not permanently changed by the reaction is called a(n)__________

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1 answer:

jekas [21]3 years ago

5 0

Answer:

catalyst

Explanation:

a substance which alters the rate of a chemical reaction but is not permanently changed by the reaction is called a catalyst

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PLS HELP ASAP!! I will do brainliest!!!! PLS IMMA DIE

Bess [88]

B. combination of all wavelengths of visible light

3 0

3 years ago

Read 2 more answers

What is the basic role of CO2 in photosynthesis? What is the basic role of CO2 in photosynthesis? CO2 is taken in by plants as a

ladessa [460]

Answer:

carbon dioxide is fixed or incorporated into organic molecules.

Explanation:

Carbon dioxide fixation is the conversion of inorganic carbon into organic carbon.

In photosynthesis CO2 is converted into glucose i.e inorganic carbon (CO2) is converted into organic molecule (glucose).

4 0

3 years ago

A particular graduated cylinder contains 24.0 mL Br2(liquid). The density of bromine at 25 degrees C is 3.12g/cm^3. Note that in

motikmotik

Answer:

$2.82x10^{23}molecules$

Explanation:

Hello.

In this case, given the volume (1cm³=1mL) and density of the bromine we are to firstly compute the mass since it will allow us to compute the representative particles:

$\rho =\frac{m}{V}$

$m=\rho *V=3.12g/cm^3*24.0cm^3\\\\m=74.88gBr_2$

Next, since the mass of one mole of diatomic bromine is 159.82 g (one bromine weights 78.91), we can next compute the moles in that sample:

$n=74.88g*\frac{1mol}{159.82g} =0.469molBr_2$

Finally, via the Avogadro's number we can compute the representative particles of bromine as follows:

$particles=0.469mol*\frac{6.022x10^{23}molecules}{1mol}\\ \\2.82x10^{23}molecules$

Best regards.

4 0

3 years ago

ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion

vfiekz [6]

Answer:

Here's what I get

Explanation:

1. Write the chemical equation

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵

Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

$K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}$

3. Set up an ICE table

A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹: 0.35 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.35-x x x

4. Solve for x

$\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}$

Check for negligibility,

$\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}$

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

pH = 9.12

Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

3 0

3 years ago

Calculate the standard potential, e°, for this reaction from its equilibrium constant at 298 k. 6.47 x 10^5

Elina [12.6K]

The working equation for this is written below:

E° = 0.0592logK/n

So, we have to know the value of n first which represents the number of moles electron in the reaction. We cannot answer this because we are not given with the reaction. However, just suppose the reaction is:

Cu⁺ + 2e⁻ --> Cu

Then, n=2. Continuing,

E° = 0.0592log(6.47×10⁵)/2 =<em> 0.172 V</em>

4 0

3 years ago