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geniusboy [140]
3 years ago
7

You have 100 grams of rubidium bromide (RbBr). How many moles do you have?

Chemistry
2 answers:
Ainat [17]3 years ago
8 0

<u>Answer:</u> The moles of rubidium bromide for the given amount is 0.61 moles

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of RbBr = 100 g

Molar mass of RbBr = 165.37 g/mol

Putting values in above equation, we get:

\text{Moles of RbBr}=\frac{100g}{165.37g/mol}=0.61mol

Hence, the moles of rubidium bromide for the given amount is 0.61 moles

Margarita [4]3 years ago
6 0

Answer:

a) 0.61 mol

Explanation:

165.3718 g Rubidium Bromide = 1 mol

100 g  Rubidium Bromide =  1/165.3718  x   100

                                          =  0.61 mol

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A stock solution is 28.2 percent ammonia (NH3) by mass, and the solution has a density of 0.8990 grams per milliliter. What volu
Savatey [412]

Answer:

\boxed{\text{20.2 mL}}

Explanation:

Assume that the volume of the stock solution is 1 L.

1. Mass of  stock solution

\text{Mass} = \text{1000 mL} \times \dfrac{\text{0.8990 g}}{\text{1 mL}} = \text{899.0 g}

2.Mass of NH₃

\text{Mass of NH}_{3} = \text{899.0 g stock} \times \dfrac{\text{28.2 g NH}_{3}}{\text{100 g stock}} = \text{253.5 g NH}_{3}

3. Moles of NH₃

\text{Moles of NH}_{3} = \text{253.5 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.89 mol NH}_{3}

4. Molar concentration of stock solution

c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{14.89 mol}}{\text{1 L}} = \text{14.89 mol/L}

5. Volume of stock needed for dilution

Now that you know the concentration of the stock solution, you can use the dilution formula .

c_{1}V_{1} = c_{2}V_{2}

to calculate the volume of stock solution.

Data:

c₁ = 14.89   mol·L⁻¹; V₁ = ?

c₂ = 0.500 mol·L⁻¹; V₂ = 600 mL

Calculations:

\begin{array}{rcl}14.89V_{1} & = & 0.500 \times 600\\14.89V_{1} & = & 300\\V_{1} & = & \text{20.2 mL}\\\end{array}\\\text{You will need $\boxed{\textbf{20.2 mL}}$ of the stock solution.}

4 0
3 years ago
Lagta ha bhaijaan sa sab bahenjaan naraz haan koi bat ni kr raha​
Fiesta28 [93]

Answer:

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Explanation:

5 0
3 years ago
A 0.479 g sample of nitrogen, oxygen or neon gas occupies a volume of 265 ml at 157 kpa and 20◦c. what is the molar mass and ide
Anna11 [10]
PV = nRT
P = 157 kPa = 157 × 10³ Pa
V = 265 ml = 0.265 l
T = 20°C = 293 K
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PV•M = mRT
M = (mRT)/(PV)
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Methane burns in oxygen to produce carbon dioxide and water. Which of the following represents
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Answer:

A

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