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Aleks04 [339]
3 years ago
10

How many grams of RbOH are present in 35.0 mL of a 5.50 M RbOH solution?​

Chemistry
1 answer:
Ganezh [65]3 years ago
8 0
I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
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How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?
rewona [7]
Answer:
             7.32 g of F₂

Solution:
              The equation is as follow,

                                   2 LiI  +  F₂    →    2 LiF  +  I₂

According to equation,

           51.88 g (2 mole) of LiF is produced from  =  37.99 g (1 mole) F₂
So,
                          10 g of LiF will be produced by  =  X g of F₂

Solving for X,
                      X  =  (10 g × 37.99 g) ÷ 51.88 g

                      X  =  7.32 g of F₂
8 0
3 years ago
Is NaCi a metal or nonmetal
Kruka [31]
Hi , NaCl is basically salt , Na for sodium and Cl for chlorine , these elements make Sodium Chloride .The bond between them is ionic.
5 0
3 years ago
Read 2 more answers
About how many elements are found on the periodic table? Question 8 options:
krek1111 [17]

the answer is actually 118 look it up

3 0
3 years ago
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
max2010maxim [7]

Answer:

The answer to your question is: Initial temperature of copper = 67.1°C

Explanation:

Data

mass Copper = 248 g

volume Water = 390 ml

T1 water = 22.6°C

T2           = 39.9°C

T1 copper = ?

Specific heat water = 1 cal/g°C

Specific heat copper = 0.092 cal/g°C

Formula       copper             water

Heat is negative for copper because it releases heat

                  - mCp(T2 - T1) = mCp(T2 - T1)                  

                  - (248)(39.9 - T1) = 390 (1)((39.9 - 22.6)           Substitution

                 -9895.2 + 248T1 = 390(17.3)                             Simplification

                 -9895.2 + 248T1 = 6747

                 248 T1 = 6747 + 9895.2

                 248 T1 = 16642.2

                 T1 = 16642.2 / 248

                 T1 = 67.1 °C                                                         Result

6 0
3 years ago
Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.
Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

\%C_E = \frac{E}{T} * 100\%

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In (Ca(OH)_2)

The amount of all elements is: (i.e formula mass of (Ca(OH)_2))

T = 1 * Ca + 2 * H + 2 * O

T = 1 * 40 + 2 * 1 + 2 * 16

T = 74

The amount of calcium is: (i.e formula mass of calcium)

E = 1 * Ca

E = 1 * 40

E = 40

So, the percentage component of calcium is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{40}{74} * 100\%

\%C_E = \frac{4000}{74}\%

\%C_E = 54.05\%

The amount of hydrogen is:

E = 2 * H

E = 2 * 1

E = 2

So, the percentage component of hydrogen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{2}{74} * 100\%

\%C_E = \frac{200}{74}\%

\%C_E = 2.70\%

Similarly, for oxygen:

The amount of oxygen is:

E = 2 * O

E = 2 * 16

E = 32

So, the percentage component of oxygen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{32}{74} * 100\%

\%C_E = \frac{3200}{74}\%

\%C_E = 43.24\%

5 0
2 years ago
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