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Inessa [10]
3 years ago
9

. In what direction is heat flowing when inserting your cold hand into a warm glove on a winter day? From the warm glove to your

cold hand From the sun to your cold hand From your cold hand to the warm glove From your warm hand to the cold glove
Chemistry
2 answers:
timofeeve [1]3 years ago
7 0

Answer:

from the warm glove, to your warm hand <3

Explanation:

stealth61 [152]3 years ago
4 0

Answer:

From the warm glove to your cold hand

Heat from the glove will be transferred to your cold hand which had less heat. Heat does NOT like gradients, so it will flow from where there is more heat to where there is less until the heat across the surface is even.

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How many significant figures are in 11.005 g​
Annette [7]

Answer:

Number of Significant Figures: 5

The Significant Figures are 1 1 0 0 5

Explanation:

8 0
4 years ago
2. A solution is made by adding 1.23 mol of KCl to 1000.0 g of water. Assume that the
ladessa [460]

Answer:

a. 74.55 g/mol.

b. 91.70 g.

c. 8.40%.

d. 1.23 mol/L.

Explanation:

<em>a. Calculate the formula weight of KCl.</em>

∵ Formula weight of KCl = atomic weight of K + atomic weight of Cl

atomic weight of K = 39.098 g/mol, atomic weight of Cl = 35.45 g/mol.

∴ Formula weight of KCl = atomic weight of K + atomic weight of Cl = 39.098 g/mol + 35.45 g/mol = 74.548 g/mol ≅ 74.55 g/mol.

<em>b. Calculate the mass of KCl in grams.</em>

  • we can use the relation:

<em>no. of moles (n) = mass/molar mass.</em>

∴ mass of KCl = n*molar mass = (1.23 mol)*(74.55 g/mol) = 91.69 g ≅ 91.70 g.

<em>c. Calculate the percent by mass of KCl in this solution.</em>

The mass % of KCl = (mass of KCl/mass of the solution) * 100.

mass of KCl = 91.70 g,

mass of the solution = 1000.0 g of water + 91.70 g of KCl = 1091.70 g.

∴ The mass % of KCl = (91.70 g/1091.70 g)*100 = 8.399% ≅ 8.40%.

<em>d. Calculate the molarity of the solution.</em>

Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(Volume of the solution (L))</em>

no. of moles of KCl = 1.23 mol,

Volume of the solution = mass of water / density of water = (1000.0 g)/(1.00 g/mL) = 1000.0 mL = 1.0 L.

M = (1.23 mol)/(1.0 L) = 1.23 mol/L.

3 0
4 years ago
Manuel is heating one liter of water to its boiling point. If one liter of water starts to boil at 100 °C, at what temperature w
Goryan [66]

C im pretty sure because its twice the amount

4 0
3 years ago
What is the linkage that binds one monosaccharide to another?
ra1l [238]

Answer:

hope this helps

Explanation:

glycosidic bond

A covalent bond formed between a carbohydrate molecule and another molecule (in this case, between two monosaccharides) is known as a glycosidic bond (Figure 4). Glycosidic bonds (also called glycosidic linkages) can be of the alpha or the beta type.

4 0
3 years ago
Read 2 more answers
An antacid tablet weighing 0.853g contained calcium carbonate as the active ingredient, in addition to an inert binder. When an
ArbitrLikvidat [17]

Answer:

0.22 g of CO2 were produced.

Explanation:

First, let's represent what is happening with an hypothetical chemical equation just to have a clearer vision of the presented process:

CaCO3 (aq) + 2 HAc (aq) → CaAc2 (aq) + H2O (l] + CO2 (g)

We have a tablet that has CaCO3 as the active ingredient that when combined with an acid, in this case represented as HAc, reacts producing a Calcium salt, water and carbon dioxide that will leave the solution as gas.

Having said that, we know that the initial mass of the reactants will have to maintain during the chemical reaction, or what is the same, the quantity of matter during the process will not change. So, if we have a tablet that weighs 0.853 g and we add an acid solution of 56.519 g, then we have that the initial mass of the reactants will be:

0.853 g from tablet + 56.519 g from acid solution = 57.372 g

This amount of matter should be the same after the reaction, but we know that the CO2 gas will leave the solution once it's formed, so considering that the resulting solution weighs 57.152 g we could calculate the mass of CO2 produced:

57.372 g of initial mass - 57.152 g of resulting solution = 0.22 g of CO2 that left the aqueous solution as gas.

5 0
3 years ago
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