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Lesechka [4]
3 years ago
9

What all indicates a CHEMICAL change? *

Chemistry
1 answer:
myrzilka [38]3 years ago
6 0

one substance becomes two new substances

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I need the answer to 452 of arogon
yaroslaw [1]
I'm confused... is there more info? 
3 0
3 years ago
use the internet to look up the sds for 2.0 m sodium hydroxide naoh to answer the following question a) list the potential acute
VashaNatasha [74]

A safety data sheet is informational data that states the properties of the various chemicals. According to the SDS of NaOH, it causes irritation, nausea, eye burn, burns in the digestive system, etc.

<h3>What is SDS?</h3>

SDS is abbreviated for a safety data sheet which is a document that provides information on the chemicals including the properties, health hazards, measures, safety, prevention, etc.

According to SDS, NaOH results in various potential acute health effects including vomiting, diarrhea, nausea, irritation, etc. The chronic effect includes coma, burns in the eye, respiratory infection, digestive system, dermatitis, etc.

Therefore, sodium hydroxide has various acute and chronic health effects.

Learn more about SDS, here:

brainly.com/question/2471127

#SPJ4

Your question is incomplete, but most probably your full question was, Use the internet to look up the sds for 2.0 m sodium hydroxide, NaOH, to answer the following question: list the potential acute and chronic health effects.

6 0
1 year ago
4.60 mL of 0.1852 M HNO3 is titrated to the phenolphthalein indicator endpoint with 27.35 mL of a KOH solution. What is the mola
Kamila [148]

Answer:

M_{base}=0.0311M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:

n_{acid}=n_{base}

Which in terms of molarities and volumes is:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the base (KOH) to obtain:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{4.60mL*0.1852M}{27.35mL}\\\\M_{base}=0.0311M

Regards!

4 0
3 years ago
Please please help me with this chemistry question over limiting reactants it’s due by midnight! I’ll mark you BRAINIEST!!!
OverLord2011 [107]

Answer: 1.5 moles

Explanation: one mole Zn uses 2 moles HCl.

1.5 moles Zn uses 3.0 mol HCl. Then Zn is a limiting reactant

And produces equal amount of H2.

4 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}

Reemplazando en la segunda ecuación:

120=\frac{90-2*a}{2} *3+\frac{9}{2} *a

Resolviendo:

120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a

120=135-3*a +\frac{9}{2} *a

120-135=-3*a +\frac{9}{2} *a

-15=\frac{3}{2} *a

\frac{-15}{\frac{3}{2} } =a

-10=a

Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

v0= \frac{90-2*(-10)}{2}

Resolviendo:

v0= \frac{90+20}{2}

v0= \frac{110}{2}

v0=55

<u><em>La velocidad inicial es 55 </em></u>\frac{m}{s}<u><em>y su aceleración es -10 </em></u>\frac{m}{s^{2} }<u><em></em></u>

3 0
3 years ago
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