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3 years ago

I’m so confused on this one!

Chemistry

1 answer:

stellarik [79]3 years ago

5 0

The correct answer is 1) Ag, Si, I₂

Ag or silver is a metal. Silver is a transition element which lies in the d-block of the periodic table.

Si or silicon is a metalloid. Silicon is a metalloid which lies in the p-block of the periodic table.

I₂ or iodine is a nonmetal. Iodine is a halogen which lies in group 17 of the periodic table. All halogens are nonmetals.

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A star is mainly made of hydrogen gas, but most classes of stars include other elements such as helium, calcium, and other heavi

Xelga [282]

That would be true, since most other stats include helium, calcium, etc

6 0

2 years ago

Read 2 more answers

How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)

Masteriza [31]

Answer: The amount of heat required to warm given amount of water is 470.9 kJ

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g$

To calculate the heat absorbed by the water, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ$

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0

3 years ago

3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on

AfilCa [17]

Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

Explanation:

Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.

3 0

3 years ago

When a radioactive isotope undergoes Beta Decay, which of the following will NOT occur? PLEASE HELP ASAP! THANKS.

sleet_krkn [62]

D) Energy is released

3 0

3 years ago

Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in

faust18 [17]

Answer:

La velocidad inicial es 55 $\frac{m}{s}$ y su aceleración es -10 $\frac{m}{s^{2} }$

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

$x=x0+v0*t+\frac{1}{2} *a*t^{2}$

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

x= 90 m
x0= 0 m
v0= ?
t= 2
a= ?

Reemplazando obtenes:

$90=v0*2+\frac{1}{2} *a*2^{2}$

$90=v0*2+\frac{1}{2} *a*4$

$90=v0*2+2*a$

Y por otro lado tenes:

x= 120 m
x0= 0
v0= ?
t= 3
a= ?

Reemplazando obtenes:

$120=v0*3+\frac{1}{2} *a*3^{2}$

$120=v0*3+\frac{1}{2} *a*9$

$120=v0*3+\frac{9}{2} *a$

Por lo que tenes el siguiente sistema de ecuaciones:

$\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.$

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

$v0= \frac{90-2*a}{2}$