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2 years ago

How closely a measurement reflects the actual value is precision. True or False??

Chemistry

1 answer:

Rina8888 [55]2 years ago

7 0

Answer:

very very very verytrue

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How many moles of Au2S3 is required to form 56 grams of H2S at STP?

Law Incorporation [45]

Answer:

0.55 mol Au₂S₃

Explanation:

Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.

1. Gather all the information in one place:

M_r: 34.08

Au₂S₃ + … ⟶ 3H₂S + …

m/g: 56

2. Calculate the moles of H₂S

Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)

= 1.64 mol H₂S

3. Calculate the moles of Au₂S₃

The molar ratio is 1 mol Au₂S₃/3 mol H₂S.

Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)

= 0.55 mol Au₂S₃

7 0

3 years ago

Consider the following statements.

Vsevolod [243]

Answer: B and C only

Explanation:

3 0

2 years ago

Papi munyanyo eeeeeeeeeeeeeeeeeeeeeeeeeeee

xenn [34]

Answer:

YES YES YES YES YES YES YES YES YES YES YES YES

8 0

2 years ago

Read 2 more answers

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

c. Solution C : $[H_3O^+]=5.65\times 10^{-4}M$

$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

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2 years ago

What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL

Rasek [7]

Density is weight by volume.

First. If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

$\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL$

1L=1000ml

$\frac{1L}{1000mL}$

$(382.14mL)( \frac{1L}{1000mL})= \frac{(382.14mL)(L)}{1000mL} =\frac{382.14L}{1000}=0.38214L$

0.38214 Liters.

4 0

3 years ago