Balance the following reaction: NaOH + HCl → NaCl + H2O Enter your answer in

100 POINTS PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

dimulka [17.4K]

Answer:

i need this im struggling with my work

Explanation:

3 0

3 years ago

Read 2 more answers

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

34kurt

Answer:

81 °C

Explanation:

This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:

q = heat added/released by a sample

m = mass of sample

c=specific heat of sample

ΔT = change in temperature

from here we can rearrange the equation to state:

q/(mc) = ΔT

1200J/((20.0g)(4.2J/g•°C)) = ΔT

14°C = ΔT

If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.

4 0

3 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago

If 12g nitrogen gas,0.40 of H2 gas and 9.0 gram of oxygen are put into 1 litre container of. 27°C what is the total pressure on

Vadim26 [7]

Answer:

Total pressure = 27.35 atm

Explanation:

Given data:

Mass of nitrogen = 12 g

Mass of H₂ = 0.40 mol

Mass of oxygen = 9.0 g

Volume of Container = 1 L

Temperature = 27 °C (27+273 = 300 K)

Total Pressure = ?

Solution:

First of all we will calculate the number of moles of individual gas.

Number of moles = mass/ molar mass

Number of moles = 12 g/ 28 g/mol

Number of moles = 0.43 mol

Pressure of N₂:

PV = nRT

P = nRT/V

P = 0.43 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 10.6 atm

Number of moles of Oxygen:

Number of moles = mass/ molar mass

Number of moles = 9 g/ 32 g/mol

Number of moles = 0.28 mol

Pressure of O₂:

PV = nRT

P = nRT/V

P = 0.28 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 6.9 atm

Pressure of H₂:

PV = nRT

P = nRT/V

P = 0.40 mol × 0.0821 atm.L/mol.K× 300 K/ 1 L

P = 9.85 atm

Total pressure:

Total pressure = Pressure of H₂ + Pressure of O₂ + Pressure of N₂

Total pressure = 9.85 atm + 6.9 atm + 10.6 atm

Total pressure = 27.35 atm

8 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago