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2 years ago

Chemistry question please help

Chemistry

1 answer:

Luda [366]2 years ago

7 0

Answer:

a) the tendency to form hard, brittle crystals.

Explanation:

Metallic bonding occurs between metals; metals are excellent conductors of electricity, are malleable (can be hammered into thin sheets) and have luster (are shiny). Brittleness is indicative of an ionic bond or ionic compound.

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Si Units Worksheet. Use si units to complete

Keith_Richards [23]

$\\ \bull\tt\dashrightarrow 2000m=2km$

$\\ \bull\tt\dashrightarrow 9000mL=9L$

$\\ \bull\tt\dashrightarrow 6L=6000mL$

$\\ \bull\tt\dashrightarrow 6cm=60mm$

$\\ \bull\tt\dashrightarrow 1000m=1km$

$\\ \bull\tt\dashrightarrow 50mm=5cm$

$\\ \bull\tt\dashrightarrow 4L=4000mL$

$\\ \bull\tt\dashrightarrow 10000g=10kg$

$\\ \bull\tt\dashrightarrow 10000m=10km$

$\\ \bull\tt\dashrightarrow 4000g=4kg$

$\\ \bull\tt\dashrightarrow 9km=9000m$

$\\ \bull\tt\dashrightarrow 3kg=3000g$

$\\ \bull\tt\dashrightarrow 90mm=9cm$

$\\ \bull\tt\dashrightarrow 4km=4000m$

$\\ \bull\tt\dashrightarrow 2000g=2kg$

$\\ \bull\tt\dashrightarrow 400cm=4m$

$\\ \bull\tt\dashrightarrow 3km=3000m$

$\\ \bull\tt\dashrightarrow 2cm=20mm$

$\\ \bull\tt\dashrightarrow 9000g=9kg$

8 0

2 years ago

A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r

slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

8 0

3 years ago

What if the mother is type O and the father is A? What would the offspring's blood type be?

tensa zangetsu [6.8K]

Answer:

it depends upon a his body which antigen is present

Explanation:

it depends on that person not on their parents

7 0

2 years ago

The high polarity of the oxygen-carbon bond in alcohols is what allows them to be soluble in water.

Anna11 [10]

This is false. An alcohol does indeed have a polar C-O single bond, but what we should really be focusing on is the extraordinarily polar O-H single bond. When oxygen, fluorine, or nitrogen is bound to a hydrogen atom, there is a small (but not negligible) charge separation, where the eletronegative N, O, or F has a partial negative charge, and the H has a partial positive charge. Water has two O-H single bonds in it (structure is H-O-H). The partially negative charge on the O of the water molecule (specifically around the lone pair) can become attracted either a neighboring water molecule's partially positive H atom, or an alcohol's partially positive H atom. This is weak (and partially covalent) attraction is called a hydrogen bond. This is stronger than a typical dipole-dipole attraction (as would be seen between neighboring C-O single bonds), and much stronger than dispersion forces (between any two atoms). When the solvent (water) and the solute (the alcohol) both exhibit similar intermolecular forces (hydrogen bonding being the most important in this case), they can mix completely in all proportions (i.e. they are miscible) in water.

8 0

3 years ago

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

8 0

2 years ago