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bagirrra123 [75]
3 years ago
13

HELP LAST QUESTION WILL MAKE BRAINLIEST

Chemistry
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

I think it is C.)????

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A student wants to make a 5.00%solution of rubidium chloride using 0.377g of the substance. what mass of water will be needed to
Alenkasestr [34]
 The mass of water that will  be needed to make the solution  is calculated as below

%  solution  =  mass of the solute/mass of the solvent(water) x100

%  solution = 5% = 5/100
mass of the solute =0.377 g
mass of the  solvent = ?

let the mass  of the solvent be  represented by Y

= 5/100 =0.377/y

by cross multiplication

5y=  37.7
divide both side by 5

y =7.54  grams
7 0
3 years ago
Helppppppppppppppppppppp
grandymaker [24]

Answer:

Formation of sedimentary rocks

Explanation:

I believe this is correct, May I get a brainliest???

3 0
3 years ago
What type(s) of atoms make up the element carbon?
ivolga24 [154]
A, as carbon is an atomic substance - element
8 0
3 years ago
Read 2 more answers
PLEASE HELP
swat32

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

5 0
3 years ago
How many moles in 4.93 x 10E23 atoms of silver?
leva [86]
<h3>Answer:</h3>

0.819 mol Ag

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})
  2. Divide:                              \displaystyle 0.818665 \ mol \ Ag

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.818665 mol Ag ≈ 0.819 mol Ag

8 0
3 years ago
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