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tresset_1 [31]
2 years ago
6

How would a fast-flowing river be most likely to move sand-sized particles of sediment?

Chemistry
1 answer:
Nitella [24]2 years ago
3 0
A fast-flowing river would most likely be to move the skaters glide around the rink b<span>y lifting it up and carrying it downstream. In a fast-flowing river, the pressure is strong, which makes it easier for sand-sized particles of sediments to be carried or transferred. Hope this answer helps.</span>
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write the steps you would use for making tea.use the words solution,solvent,solute,dissolve,soluble,insoluble,filtrate and resid
poizon [28]
1. Take 100ml of water as solvent and boil it few minutes.
2. Now add one tea spoon sugar, one tea spoon tea leaves and 50ml of milk. Here sugar, tea leaves and milk are solute.
3. Now boil it again for few minutes so that sugar will dissolves in solution as sugar is soluble in water
4. Now filter the solution. Collect the filtrate in cup. The insoluble tea leaves will be left behind as residue.
4 0
2 years ago
Please help me with this question!! read the photo
pishuonlain [190]

There are 1.92 × 10^23 atoms Mo in the cylinder.

<em>Step 1</em>. Calculate the <em>mass of the cylinder </em>

Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g

<em>Step 2</em>. Calculate the<em> mass of Mo </em>

Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo

<em>Step 3</em>. Convert <em>grams of Mo</em> to <em>moles of Mo </em>

Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo

<em>Step 4</em>. Convert <em>moles of M</em>o to <em>atoms of Mo </em>

Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^2<em>3</em> atoms Mo)/(1 mol Mo)

= 1.92 × 10^23 atoms Mo

7 0
2 years ago
No caso da pessoa ser um atleta profissional,essa demanda de energia pode subir até 5400Kcal. Transforme essa medida de energia
ddd [48]

Answer:

If you speak Any English i Think I will be able to help you Los Amigo

Explanation:

6 0
2 years ago
Consider the following scenario
GarryVolchara [31]

Answer:

See explaination

Explanation:

Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,

AgF + NaCl= AgCl + NaF

Here, the color of AgCl is white, hence the solution cannot be AgCl.

Determination of NaCl

Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.

Reactions

Ag+ (aq)+ Cl-(aq) = AgCl(aq)

Ag+(aq) + SCN-(aq) = AgSCN(aq)

Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)

7 0
3 years ago
Draw a lewis structure for ketene, c2h2o, which has a carbon–carbon double bond
MrMuchimi

Visual representation of covalent bonding indicating the valence shell electrons in the molecule, lines represents the shared pair of electron and pair of electrons that are not involved in bonding are represented as dots(lone pairs) are known as Lewis structures.

Compound formation takes place in order to complete the octet of each element that is according to octet rule, each atom forms bond with other atom in order to complete their octet that is to get eight electrons in its valence shell and attain stability.

An organic compound of the form R^{'}R^{''}C=C=O is known as ketene.

The given ketene is C_2H_2O.

The number of valence electron of:

C = 4

H = 1

O = 6

The number of valence electrons in C_2H_2O = 2\times 4+2\times 1+1\times 6 =16

2 electrons are involved in each single bond between carbon and hydrogen and 4 electrons are involved in each double bond formed between carbon-carbon and carbon-oxygen. Hence, the total number of electrons involved in bond formation are 12 and rest 2 pair of electrons are present on oxygen as lone pair of electrons.

Therefore, the attached image is the Lewis structure of C_2H_2O .

8 0
3 years ago
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