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nordsb [41]
3 years ago
7

The emission spectrums are known for the elements below. Line spectra for boron, lithium, hydrogen, helium, potassium and unknow

n sample. Boron has 3 purple lines, 1 blue line, 2 blue-green lines and 2 orange lines. Lithium has 2 purple lines, 2 blue lines, 2 blue-green lines, 1 orange line and 1 red line. hydrogen has one line of each purple, blue, blue-green and red. helium has 3 purple lines, 3 blue lines, 3 blue-green lines 1 yellow line and 2 red lines. Potassium has 8 purple lines, 6 blue-green lines, 10 green lines, 3 yellow lines, 2 orange lines and 6 red lines. The unknown sample has Potassium has 8 purple lines, 6 blue-green lines, 10 green lines, 3 yellow lines, 2 orange lines and 6 red lines. What elements are in the unknown sample? Check all that apply. boron lithium hydrogen helium potassium
Chemistry
2 answers:
Montano1993 [528]3 years ago
8 0

Answer:

E

Explanation:

nata0808 [166]3 years ago
7 0

Answer:

The Answer is (B) lithium and (E) potassium

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Differentiate between a precipitate and an aqueous solution
antiseptic1488 [7]

Answer:

A precipate is a solid while an aqueous solution is liquid.

Explanation:

A precipitate is a solid which separates after a chemical reaction occurs. It is the solid product of the reaction.

An aqueous solution is formed when a substance is dissolved in water.

6 0
3 years ago
I need the answer please
dlinn [17]

Answer:

Na+ the answer is A

Explanation:

6 0
3 years ago
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When the metallic element sodium combines with the nonmetallic element bromine (Br2), is the product a solid, liquid, or gas at
Luden [163]

Answer: it is a solid at room temperature.

Explanation:

The ions of the two reactants are Na{+} and Br{-}, so the formula of the product is NaBr.  

and when the NaCl is a solid a room temperature. and when NaBr has similar properties to NaCl, so NaBr is a solid at room temperature too.

so our chemical formula for the product is NaBr and It's a solid at room temperature.

4 0
2 years ago
For the reaction co(g)+2h2(g)<->ch3oh(g) at 700. k, equilibrium concentrations are [h2] = 0.072 m, [co] = 0.020 m, and [ch
SVETLANKA909090 [29]
The balanced equation for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃<span>OH(g)

Since given concentrations are at equilibrium state, the expression for the equilibrium constant, k can be written as
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   k = 0.030 M / 0.020 M x (<span>0.072 M</span>)²
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6 0
3 years ago
100 PIONTSSSSS HELP ASAP
valina [46]

Left Panel

Short answer A

<em><u>Solution</u></em>

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN  THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

<em><u>Step One</u></em>

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

<em><u>Step Two</u></em>

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

<em><u>Step Three</u></em>

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

<em><u>Step Four</u></em>

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

<em><u>Step Five</u></em>

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P  = 2 * 31 =    62

8 O = 8 * 16   =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

<em><u>Step Six</u></em>

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

<em><u>Step One</u></em>

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass =  141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass  multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C


6 0
3 years ago
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