All categories

3 years ago

On a separate sheet of paper, write the formulas for the missing components of these neutralization reactions.

Chemistry

2 answers:

Fofino [41]3 years ago

4 0

The formula of the missing component are a follows

HBr + KOH ------> KBr + H2O

H2SO4 + 2NH4OH--------> (NH4)2SO4 + 2H2O

2HNO3 + Mg(OH)2 ---------> Mg(NO3)2 + 2 H2O

PolarNik [594]3 years ago

4 0

The formula of the missing component are a follows

HBr + KOH ------>  KBr  + H2O

H2SO4 +  2NH4OH--------> (NH4)2SO4 + 2H2O

2HNO3    + Mg(OH)2 ---------> Mg(NO3)2 + 2 H2O
PLEASE RATE BRAINLIEST

You might be interested in

How many Electrons, Neutrons & Protons do

EastWind [94]

Answer:

sodium

protons:11

neutrons:12

electrons:11

nitrogen

protons:7

neutrons:7

electrons:7

6 0

3 years ago

Read 2 more answers

If I have 6 moles of a gas at a pressure of 3.4 atm and a volume of 10 liters, what

ch4aika [34]

Answer:

The temperature is 69.05 K.

Explanation:

We have,

Number of moles are 6

Pressure is 3.4 atm and a volume of 10 liters.

It is required to find the temperature. It can be calculated using gas law equation. It says that,

$PV=nRT$

P is pressure

V is volume

n is number of moles

R is gas constant, $R=0.082057\ L-atm/mol-K$

T is temperature

Plugging all the values we get :

$T=\dfrac{PV}{nR}\\\\T=\dfrac{3.4\times 10}{6\times 0.082057}\\\\T=69.05\ K$

So, the temperature is 69.05 K.

8 0

3 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$