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vovikov84 [41]
2 years ago
14

50 mL graduated cylinder contains 25.0 mL of water. A 142.5040 g piece of osmium is placed in the graduated cylinder and the wat

er level rises to 31.3 mL. What is the density of the piece of osmium?
Chemistry
1 answer:
GrogVix [38]2 years ago
8 0

Answer:

22.6g/mL

Explanation:

density = mass / volume

given mass of object : 142.5040

volume of object : not given however we are told that when placed inside of a cylinder with 25 mL of water the water level rises to 31.3 mL

31.3 - 25 = 6.3

So the water level rose by 6.3 mL meaning that the object has a volume of 6.3 mL

Now to find the density.

Recall that density = mass / volume

mass = 142.5040 g and volume = 6.3 mL

so density = 142.5040g / 6.3 mL = 22.6 g / ml

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Answer: option C) gather information and identify stakeholders

Explanation:

The sales, distribution and advertisement of alcoholic beverages requires information on consumer protection, health risks, and environmental factors. And Garrett would simply get such from the relevant stakeholders like regulatory agencies.

Thus, Garrett should first gather information and identify stakeholders

3 0
3 years ago
Write 10,847,100 in Scientific Notation with 4 significant figures.
masha68 [24]

Answer:

The number 10,847,100 in Scientific Notation is 1.0847x10^{7}

Explanation:

Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.

5 0
3 years ago
1. The multiplicative inverse of 5/9 is.....<br>a-9/5<br>b-9/5<br>c-3/9<br>d-None of these​
Luda [366]

Answer:

None of these cause the correct answer is 9/-5

5 0
3 years ago
Read 2 more answers
Calculate the mass of ammonia (NH3) that contains a billion (1.00 * 10^12) hydrogen atoms.
Lynna [10]

Answer:

The mass of ammonia (NH3) that contains 1.00 \times 10^12 hydrogen atoms is 9.4\times 10^{-12} g.

Explanation:

As 6.022\times 10^{23} atoms of hydrogen = 1 mole of the hydrogen atom

Therefore, 10^{12} atoms of hydrogen = \frac{1}{6.022 \times 10^{23}}\times 10^{12}=1.66\times 10^{-12} moles of the hydrogen atom.

Now, there are 3 moles of hydrogen atoms in 1 mole of ammonia (NH_3).

As the mass of 1 mole of ammonia is 17g, so

when there are 3 moles of hydrogen atoms, then the mass of ammonia = 17 g

Therefore, when there are 1.66\times 10^{-12} moles of hydrogen atoms, then the mass of ammonia = \frac{17}{3}\times 1.66\times 10^{-12}=9.4\times 10^{-12 g.

Hence, the mass of ammonia (NH_3) that contains 1.00 \times 10^12 hydrogen atoms is 9.4\times 10^{-12} g.

3 0
3 years ago
Benzene is a minor component of gasoline. The standard molar enthalpy of formation of benzene C7H16(l) is 48.95 kJ/mol. For the
Radda [10]

Answer:

-3135.47 kJ/mol

Explanation:

Step 1: Write the balanced equation

C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)

Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)

We will use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpies of formation

p: products

r: reactants

ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))

ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol

ΔH°r = -3135.47 kJ

Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.

7 0
2 years ago
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