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3 years ago

If y gets smaller as x gets bigger, x and y have a(n) __ relationship.

Chemistry

1 answer:

Aleksandr [31]3 years ago

0 0

Answer:

Inverse relationship with each other

Explanation:

The inverse relationship cause increasing of one quantity and decreasing in the other

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What are two examples of a chemical change

lyudmila [28]

Answer:

1.Cooking an egg

2. Burning wood

Hope it helps :)

6 0

3 years ago

Read 2 more answers

When an irregularly shaped chunk of an unknown metal with a mass of 25.32 g was placed in a graduated cylinder containing 25.00

inn [45]

Answer: The density of the unknown metal is 7.86 g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of metal = 25.32 g

Volume of metal = volume of water displaced = (28.22 - 25.00) ml = 3.22 ml

Putting in the values we get:

$density=\frac{25.32g}{3.22ml}$

$Density=7.86g/ml$

Thus the density of the unknown metal is 7.86 g/ml

7 0

3 years ago

A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water at 15.4 0C and it reach a thermal equilibrium of 17.

zysi [14]

Answer:

The initial temperature of the metal is 84.149 °C.

Explanation:

The heat lost by the metal will be equivalent to the heat gain by the water.

- (msΔT)metal = (msΔT)water

-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C

-ΔT = 836.29/12.51 °C

-ΔT = 66.89 °C

-(T final - T initial) = 66.89 °C

T initial = 66.89 °C + T final

T initial = 66.89 °C + 17.3 °C

T initial = 84.149 °C.

7 0

3 years ago

How many carbon atoms does the common group of amino acids have? a. 0 c. 2 b. 1 d. 3

zzz [600]

I definitly believe the answer is c. 2

4 0

4 years ago

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

$Cl^-\ (r^-) = 1.82\ \dot{A}$

To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {6.28}{2}$

<u>Size of the potassium ion = 1.33 Å</u>

4 0

3 years ago