State one reason for the color change in beaker A

How many significant figures are in this number? 43.55

leva [86]

Answer:

4 significant figures

Explanation:

Significant figures are the units/digits within a number that make the number more accurate and precise.

All digits (except for 0) are always significant. Therefore, all the digits in 43.55 are significant. Since there are 4 digits in the given number, there are 4 significant figures.

7 0

2 years ago

A 2.6 liter container of nitrogen had a pressure of 3.2 atm. What volume will the gas occupy at a pressure of

Dominik [7]

Answer:

A sample of oxygen gas occupies a volume of 250. ... volume will it occupy at 800. torr pressure? ... A 2.0 liter container of nitrogen had a pressure of 3.2 atm. ... A sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm producing

Explanation:

8 0

3 years ago

The molar ratio of HPO42- to H2PO4- in a solution is 1.4. Calculate the pH of the solution. Phosphoric acid (H3PO4) is a triprot

Bezzdna [24]

Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.

pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.

7 0

3 years ago

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$