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3 years ago

HELP Which of the following fractions can be used in the conversion of 32 m3 to the unit mm3?

1 answer:

7 0

We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m

We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
1 mm³ = (10⁻³)³ m³

Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³

When we multiply by this conversion, we will get:
32 m³ = 32 x 10⁹ mm³

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Using the balanced equation below,

Zielflug [23.3K]

Answer:

Mass = 6.538 g

Explanation:

Given data:

Mass of zinc hydroxide produced = 9.65 g

Mass of zinc required = ?

Solution:

Chemical equation:

Zn + 2MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃

Number of moles of zinc hydroxide:

Number of moles = mass/molar mass

Number of moles = 9.65 g/ 99.42 g/mol

Number of moles = 0.1 mol

now we will compare the moles of zinc and zinc hydroxide,

Zn(OH)₂ : Zn

1 : 1

0.1 : 0.1

Mass of zinc required:

Mass = number of moles × molar mass

Mass = 0.1 mol × 65.38 g/mol

Mass = 6.538 g

8 0

2 years ago

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How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 4

Yuki888 [10]

1) (Hvap)(moles of water)=236.9783574kJ
(40.67)(105/18.02)
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(100)(105/18.02)(75.4)/1000
3) (Hfus)(moles of water)=35.01942286kJ
(6.01)(105/18.02)
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(15)(105/18.02)(36.4)/1000
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2 years ago

5. Which of the following can be classified as a mixture ?

podryga [215]

5. B air is a mixture of many elements, but is not a chemically fused.

6. B Beef stew, composition varies throughout.

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3 years ago

Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO

sergey [27]

The molarity of a solution is the number of moles of a substance divided by the volume in liters prepared.

$molarity=\frac{n}{V}$ , where n is number of moles and V is the volume in liters.

In order to calculate the mass of solute we need to convert the volume and molarity to moles

$1.421 L solution \times\frac{0.0730 moles}{1 Lsolution}= 1.037 mol NaCO_3$

Now that we have moles we use the relative formula mass of NaCO₃, We have 1 Na atom, 1 C atom and 3 O atoms, thus

$M_r= (1\times 22.99) + (1\times 12.00) + (3\times 16.00)= 82.99g/mol$

$1.037 \times\frac{82.99g}{mol} = 86.1g$

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3 years ago

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A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c

kirill [66]

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2 years ago

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