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4 years ago

Each substance on the left side of the arrow in a chemical equation is a ____.

Chemistry

1 answer:

S_A_V [24]4 years ago

7 0

Answer:

Each substance on the left side of the arrow in a chemical equation is a reactant.

Explanation:

The chemical changes (chemical reactions) are represented by chemical equations.

The chemical equations show the starting substances and the final substances.

The starting substances of the reaction are the reactants.

A single arrow is used to indicate the direction of the reaction, the reactants are placed on the left side of the equation, and the products appear on the right side.

This sketch shows that:

reactants yield products

general form A + B → C + D

example Na + Cl → NaCl

You might be interested in

The nonvolatile, nonelectrolyte DDT, C14H9Cl5 (354.50 g/mol), is soluble in diethyl ether CH3CH2OCH2CH3. Calculate the osmotic p

aleksandrvk [35]

Answer:

2.50 atm

Explanation:

We have 10.4 g of DDT (solute), whose molar mass is 354.50 g/mol. The corresponding moles are:

10.4 g × (1 mol/354.50 g) = 0.0293 mol

The molarity of the solution is:

M = moles of solute / liters of solution

M = 0.0293 mol / 0.286 L

M = 0.102 M

We can find the osmotic pressure (π) using the following pressure.

π = M × R × T

where,

R: ideal gas constant

T: absolute temperature

π = M × R × T

π = 0.102 M × 0.0821 atm.L/mol.K × 298 K

π = 2.50 atm

5 0

3 years ago

(C=12.01 amu, H=1.008 amu, O=16.00 amu) [?] g/mol CH20

tekilochka [14]

CH₂O = C + 2.H + O

= 12.01 + 2 x 1.008 + 16

= 30.026 amu = 30.026 g/mol

8 0

2 years ago

Consider the following balanced equation:

Alenkasestr [34]

Answer:

Explanation:

bs I just done that on my work and asked my teacher if it was right she said it was

3 0

4 years ago

If air has a dry-bulb temperature of 2°C

Allisa [31]

Answer:

The relative humidity is 36%.

Explanation:

Relative humidity is the measure of ratio of actual vapor density to saturated vapor density in a given system. So there are different ways to measure the relative humidity in the atmosphere. One of the common ways is to measure using Psychrometer. In this instrument, two different thermometers are used as measuring device. One of the thermometer measures the humidity in dry air and another in wet air. So the difference of the temperature of dry and wet air bulb will be related to determine the relative humidity percentage. The formula to determine the relative humidity using this method is as below.

$Relative humidity=\frac{(e_{w}-[N*(1+0.00115*T_{w}*(T_{d}-T_{w}))])*100}{e_{d} }$

Also,

$e_{d}=6.112*e^{[\frac{(17.502*T_{d} )}{(240.97+T_{d} )}]}$

And,

$e_{w}=6.112*e^{[\frac{(17.502*T_{w} )}{(240.97+T_{w} )}]}$

Here e =2.718, $T_{d}$ and $T_{w}$ are the temperature at dry bulb and wet bulb temperatures, respectively.

So, the relative humidity after substituting the values of Td = 2 C and Tw = -2 C is 36%.

6 0

3 years ago

Una masa de aire ocupa un volumen de5litro a una temperatura de 120c Cual será el nuevo volumen si la temperatura se reduce ala

zalisa [80]

Answer:

1. $V_2=2.5L$

2. $V_2=8000mL$

3. $V_2=176.3L$

Explanation:

¡Hola!

En este caso, dada la información para estos problemas, procedemos de la siguiente manera, basado en las leyes de los gases ideales:

1. Una masa de aire ocupa un volumen de 5 litros a una temperatura de 120 °C Cual será el nuevo volumen si la temperatura se reduce a la mitad:

Aqui, utilizamos la ley de Charles, asegurándonos que la temperatura está en Kelvin:

$\frac{T_2}{V_2} =\frac{T_1}{V_1} \\\\V_2 =\frac{V_1T_2}{T_1} \\\\V_2 =\frac{5L*196.5K}{393K} \\\\V_2=2.5L$

2. Un gas ideal ocupa un volumen de 4000 ml a una presión absoluta de 1500 kilo pascal Cual será la presión si el gas es comprimido lentamente hasta 750 kilo pascal a temperatura constante?

Aquí, utilizamos la ley de Boyle, dado que la temperatura se mantiene constante, calculando el volumen, ya que lo que se da es la presión final:

$\neq P_2V_2=P_1V_1\\\\V_2=\frac{P_1V_1}{P_2}\\\\ V_2=\frac{4000mL*1500kPa}{750kPa}\\\\V_2=8000mL$

3. Un gas ocupa un volumen de 200 litros a 95°C y 782 mmHg Cual será el volumen ocupado por dicho gas a 65°C y 815 mmHg:

Aquí, utilizamos la ley combinada de los gases ideales, asegurándonos que las temperaturas están en Kelvin:

$\frac{T_2}{P_2V_2} =\frac{T_1}{P_1V_1} \\\\V_2 =\frac{P_1V_1T_2}{P_2T_1} \\\\V_2 =\frac{782mmHg*200L*338K}{815mmHg*368K}\\\\V_2=176.3L$

¡Saludos!

6 0

3 years ago