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BartSMP [9]
4 years ago
5

Which of the following equations can be used to solve volume of a solution using

Chemistry
1 answer:
Rudik [331]4 years ago
7 0

Answer:Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.

Explanation: c1V1 = c2V2

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If 0.8 g of gas Z occupies 0.56 dm at s.1.p., calculate its relative molecular mass. (Molar volume of gas is 22.4 dm.)​
photoshop1234 [79]

Answer:

32

Explanation:

0.56dm of gas has 0.8g

22.4dm of gas Z has a mass of 22.4/0.56X0.8

the relative molecular mass of gas Z is 32

8 0
3 years ago
Consider a hypothetical reaction in which A and B are reactants, and C and D are products. If 26 grams of A completely reacts wi
natali 33 [55]

Answer:

The answer to your question is: 24 grams of D

Explanation:

To answer this question we need to remember the Lavoisier law of conservation of mass, which says that in a chemical reaction matter is neither created nor destroyed.

This means that the amount of matter stays the same.

Then, the reaction is

                    A      +      B      ⇒       C      +      D

                 26 g         12 g             14 g            x

mass                              

of reactants      38 g                              ?          mass of products, but it must be

                                                                          equal to the mass of products

Then                14g + x = 38

                        x = 38 - 14

                        x = 24 g of D                                                        

4 0
3 years ago
(55 points) Please help me with my homework please please help me thanks so much
balandron [24]

Answer:

I didn't do the observation so I can't help sorry

8 0
3 years ago
Which is the molar mass of BF3?
Eduardwww [97]
The answer is 67.82 g/mol
3 0
4 years ago
At its critical point, ammonia has a density of 0.235 g cm23. You have a special thick­walled glass tube that has a 10.0­mm outs
salantis [7]

Answer:

\large \boxed{\text{69.3 mg}}

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}

4 0
3 years ago
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