Cuales son los nombres de los siguientes iones NH4+1 Y NO3-1

Pls help me here, thanks in advance

Marat540 [252]

Answer:

What's the problem???

4 0

3 years ago

Read 2 more answers

`I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon i

Andrew [12]

Answer:

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law

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

Given ,

V₁ = 439 mL = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

$\frac{0.439}{317.15}=\frac{0.378}{T_2}$

$T_2=\frac{0.378\cdot \:317.15}{0.439}=273.08\ K$

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15

So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>

3 0

3 years ago

What would cause the pressure of a gas to increase?

IrinaK [193]

Answer: The pressure of a gas will increase when there is a decrease in the volume of the gas.

Explanation: according to Boyle's law, the volume of a gas will decrease when the pressure is increased at constant temperature and vice versa.

8 0

3 years ago

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

4 0

3 years ago

Plete the following problems significant figures: 6 × 0.30<br>

jeka57 [31]

6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.

0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.

3 0

3 years ago