An element's atomic number is 87. how many protons would an atom of this element have

What volume of 0.100 M HCl will react completely with 50.00 mL of 0.200 M NaOH?

skelet666 [1.2K]

Equation of reaction
Hcl+NaoH-->Nacl+H2O
1:1
Using the formula CaVa/CbVb=na/nb
Ca(Concentation of acid)= 0.100M
Cb(Concentration of base)=0.200M
Va=?
VB=50.00ml
na=1
nb=1
from the formula
Va= CbVbna/Canb
Va= 0.2*50*1/0.1*1
=10/0.1=100ml
Therefore volume of acid = 100ml

7 0

3 years ago

Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.

attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

<em>Equivalentes KOH:</em>

0.100L * (1eq / L) = 0.100eq = 0.100moles

<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>

8 0

3 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

3 years ago

Someone answer? You don’t have to answer both if you don’t know.

natima [27]

I’m not sure what the 6 means, but for the first one, the process being shown is photosynthesis that occurs in a plant. The first two compounds are the reactants. The CO2 is carbon dioxide, and the H2O is water. When these two reactants are put together with some sunlight, they create the products. The C6H12O6 is glucose, and the O2 is oxygen.

4 0

3 years ago

You are making homemade ice cream and can use either 200 g Ice Melt (CaCl2) or 200 g rock salt (NaCl) to lower the freezing poin

mash [69]

Answer:

Calcium chloride (CaCl2) would help me eat my ice cream faster

Explanation:

Addition of salt to ice melts the lowers the freezing temperature of the ice thus melting the ice easily

Adding calcium chloride to my ice cream would help me eat it faster because the melting point of calcium chloride (772°C) is lower than that of sodium chloride (melting point is 801°C). Calcium chloride would melt my ice cream faster because of its lower melting point

5 0

3 years ago