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3 years ago

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Use the mass and volume data to calculate the density of mercury to the nearest tenth. Mass of mercury = 57 g Volume of mercury

= 4.2 mL

2 answers:

Elena-2011 [213]3 years ago

7 0

Density = mass/volume = 57g/4.2 mL ≈ 13.6 g/mL

hodyreva [135]3 years ago

7 0

Use the mass and volume data to calculate the density of mercury to the nearest tenth.

Mass = 57 g Volume = 4.2 mL

The density of mercury is ⇒ 13.6 g/cm3.

Trust me i got it wrong and it gave me this i just did the instruction

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A liquid has a density of 0.87g/ml. What volume is occupied by 25g of liquid

irga5000 [103]

Answer:

29 mL

Explanation:

<h3>Equation</h3>

The question needs us to find the volume of the liquid. The equation for volume using density and mass is:

Volume = Mass / Density

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We can substitute the given values for density and mass into the equation:

$V=\frac{25\ g}{0.87\ g/ml}$

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<h3>Additional Comments</h3>

The answer we obtained (29 mL) is rounded to two significant figures. When multiplying or dividing, the amount of significant figures in the final answer is always the least amount of significant figures in one of the values.

Below are the significant figure rules:

Nonzero digits will always be significant (eg. 54 --> 2 significant figures)

Zeroes at the beginning of a number will never be significant (eg. 0.1 --> 1 significant figure)

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5 0

1 year ago

Which of these takes place when a chemical change occurs?

zmey [24]

Answer:

In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change.

Explanation:

7 0

3 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

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The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

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Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

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2 years ago

a mixture of carbon and sulfur has a mass of 9.0 g. complete combustion with excess o2 gives 22.5 g of a mixture of co2 and so2.

photoshop1234 [79]

S + O2 → SO2
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<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>

7 0

3 years ago

Use the equation n2 + 3h2=2nh3 if 8.0g N2 react, how many grams of NH3 will be produced

Eduardwww [97]

the balanced equation for the formation of ammonia is

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according to the molar ratio,

1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant

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6 0

3 years ago