All categories

2 years ago

(-5)+(+7)-(-4) + (+12)A 8B 10C 18d 20

Chemistry

1 answer:

Vladimir79 [104]2 years ago

5 0

Answer:

C-18

Explanation:

Step one follow order of operations

Add and subtract from left to right(-5)+(7)=2-(-4)+(12)

STEP 2

Apply negative Rule -(-4)=+4=2+4+(12)

then add 2+4+12=18

You might be interested in

When does the object change direction?

Tatiana [17]

Answer:

50distance and 10minutes time

Explanation:

because it's the best object

4 0

2 years ago

What is the color of the indicator thymol blue in a solution that has a ph of 11?.

Leviafan [203]

Answer:

Red

Explanation:

Red color is evidenced when thymol blue indicator is in a solution having a pH of 11. A pH of 11 means that the solution is basic or alkaline. Therefore, the indicator turns red, indicating that the solution is alkaline.

When thymol blue indicator is in a acidic solution, the indicator remains blue.

8 0

2 years ago

Are metallic bonds are found in compounds

Ratling [72]

Answer:

Explanation:

Yes :)

Mark brainliest please

5 0

2 years ago

Read 2 more answers

Identify the cfc with the shortest atmospheric lifetime

goldenfox [79]

CFC-11 is typically the CFC with the shortest atmospheric lifetime.

5 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

2 years ago

(-5)+(+7)-(-4) + (+12)A 8B 10C 18d 20​

(-5)+(+7)-(-4) + (+12)A 8B 10C 18d 20