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Aloiza [94]
2 years ago
5

Please help noowwwww... if you don’t know please don’t answer

Chemistry
1 answer:
pshichka [43]2 years ago
7 0

Answer:

i honetly dont thing anyone knows that so look it up

Explanation:

You might be interested in
How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----> K3PO4 + 3
8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

\frac{3}{1} =\frac{y}{0.02256} \\

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

5 0
3 years ago
Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which
prohojiy [21]

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

<u>Step 1:</u> Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

<u>Step 2:</u> Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

<u>Step 3:</u> Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

8 0
2 years ago
The incredible catalytic power of enzymes can perhaps best be appreciated by imagining how challenging life would be without jus
ioda

Answer:

t = 7.58 * 10¹⁹ seconds

Explanation:

First order rate constant is given as,

k =  (2.303 /t) log  [A₀] /[Aₙ]

where  [A₀]  is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>

[A₀]  = 615 calories;

[Aₙ] = 615 - 480 = 135 calories

k = 2.00 * 10⁻²⁰ sec⁻¹

substituting the values in the equation of the rate constant;

2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)

(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)

t = 2.303 / 3.037 * 10⁻²⁰

t = 7.58 * 10¹⁹ seconds

8 0
2 years ago
Chapter 7 Chemistry #52-71
Sophie [7]

If you want more instant and correct answers from people in brainly in the future please provide more information to the question you ask please and thank you. -Rogue

6 0
3 years ago
What volume, in milliliters, of a 0.997 M KOH solution is needed to neutralize 30.0 mL of 0.0400 M HCl?
deff fn [24]

Answer:

1.2 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

HCl + KOH —> KCl + H₂O

From the balanced equation,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of base, KOH (n₆) = 1

Finally, we shall determine the volume of the base, KOH needed to neutralize the acid, HCl as follow:

Molarity of base, KOH (M₆) = 0.997 M

Volume of acid, HCl (Vₐ) = 30 mL

Molarity of acid, HCl (Mₐ) = 0.0400 M

Volume of base, KOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.04 × 30 / 0.997 × V₆ = 1/1

1.2 / 0.997 × V₆ = 1

Cross multiply

0.997 × V₆ = 1.2

Divide both side by 0.997

V₆ = 1.2 / 0.997

V₆ = 1.2 mL

Thus, the volume of the base, KOH needed to neutralize the acid is 1.2 mL.

7 0
3 years ago
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