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LiRa [457]
3 years ago
11

Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B How many boxes of B would be required to make 30

grams of C?
Chemistry
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer: Two boxes of B would be required.

Explanation: From given information, A = 10 grams and B = 5 grams

C is made up of one A and one B box.

So, mass of C = mass of A + mass of B

mass of C = 10 g + 5 g = 15 g

number of boxes of C in 30 grams will be = 30g(\frac{1box}{15g})

= 2 boxes

So, if we want to make the two boxes of C then we also have to double the number of boxes of A and B.

Hence, to make 30 grams of C, we need to take two boxes of B since each box of C requires one box of A and one box of B.

Iteru [2.4K]3 years ago
5 0
Mass box C is 10+5. (So C is 15)

But if C was 30, how many times could you put B (5) into it?

30/5 = 6

You would need 6 boxes of B to make 30 grams of C.
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Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =
julia-pushkina [17]

The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

  • Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
  • Density of water = 1 g/mL
  • Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

Mass of water = 400 g

Finally, we shall determine the mole of the water

  • Mass of water = 400 g
  • Molar mass of water = 18.02 g/mol
  • Mole of water = ?

Mole = mass / molar mass

Mole of water = 400 / 18.02

Mole of water = 22.2 moles

<h3>How to de terminethe mole of KBr</h3>
  • Mass of KBr = 0.3 g
  • Molar mass of KBr = 119 g/mol
  • Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>
  • Mole of KBr = 0.0025 mole
  • Mole of water = 22.2 moles
  • Total mole = 0.0025 + 22.2 = 22.2025 moles
  • Mole fraction of KBr =?

Mole fraction = mole / total mole

Mole fraction of KBr = 0.0025 / 22.2025

Mole fraction of KBr = 0.0001

Learn more about mole fraction:

brainly.com/question/2769009

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6 0
2 years ago
This is the balances equations. C3H8 + 5O2 → 3CO2 + 4H2O How many moles of oxygen are required to produce 37.15 g CO2
AleksandrR [38]
Molar mass O2 = 31.99 g/mol

Molar mass CO2 = 44.01 g/mol

Moles ratio:

<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O 
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2

mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01

mass of O2 = 8174.8575 / 132.03

mass of O2 = 61.916 g 

Therefore:

1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916

moles O2 = 61.916 x 1 / 31.99

moles = 61.916 / 31.99 => 1.935 moles of O2
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4 years ago
Which phrase best discribes wave motion
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Answer:

propagation of disturbances—that is, deviations from a state of rest or equilibrium—from place to place in a regular and organized way. Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties.

Explanation:

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3 years ago
Is Columns of elements in the periodic table are called periods correct in the underlined sentence
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I don't know sorry I really need point
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Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n
olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

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The ionic equation in separated aqueous solution will be,

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In this equation, Ca^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

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