Dont need work pls help

Which statement describes the periodic law?

mr_godi [17]

Answer is: elements in the same group have similar chemical properties.
Periodic law<span> is the arrangement of the </span>elements<span> in order of increasing atomic number.
</span>For example all alkaline metals (I group of periodic table, Na, K, Cs...) loose one electron in chemical reaction and react vigorously with water.

3 0

3 years ago

Read 2 more answers

What was one idea Dalton taught about atoms?

gizmo_the_mogwai [7]

Explanation:

All atoms of one type were identical in mass and properties.

3 0

3 years ago

A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express

Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

(c) Molarity = 13792 mol HNO3/m3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

$wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63$

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

$V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w$

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

$\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3$

(c) HNO3 molarity (mol HNO3/m3)

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

$Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3$

We can now calculate the molarity as

$Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}$

8 0

3 years ago

Citric acid is often used as an acidity regulator in hot water canning of tomatoes. At 25°C it has Kat 7.4x10- and A.Hº = +4.1 k

gavmur [86]

Answer:

The rate constant at T = 100 C is 1.0*10⁻³

Explanation:

The Arrhenius equation relates two rate constants K1 and K2 measured at temperatures T1 and T2 as shown below:

$ln\frac{K_{2}}{K_{1}}=\frac{\Delta H^{0}rxn}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

here, ΔHrxn = standard enthalpy change of the reaction

R = gas constant

From the given information:

K1 = 7.4*10^-4

T1 = 25 C = 25+273 = 298 K

T2 = 100 C = 100+273 = 373K

ΔH°=4.1kJ/mol

$ln\frac{K_{2}}{7.4*10^{-4}}=\frac{4.1 kJ/mol}{0.08314kJ/mol.K}(\frac{1}{298}-\frac{1}{373})K$

K2 = 1.03*10⁻³

6 0

3 years ago

Which subatomic particle is electrically neutral and found in the nucleus of an atom?

tangare [24]

The neutron is electrically neutral.

5 0

3 years ago