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Cloud [144]
1 year ago
7

Convert 1.21 kg to grams

Chemistry
2 answers:
MissTica1 year ago
7 0

Answer:

1210 grams

Explanation:

Message me for extra info.

snap- parkguy786

lana66690 [7]1 year ago
5 0

Answer:

1210

Explanation:

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A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,
Alja [10]

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

-Q= 66,300 J

-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

 Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J                117 J/0C x

------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

3 0
1 year ago
Write an equation that represents the action in water of rubidium hydroxide as an Arrhenius base.
Anika [276]

Answer:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

Explanation:

Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.

Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

3 0
1 year ago
create a flowchart that outlines when to use dimensional analysis and when to use scientific notation.
murzikaleks [220]

Answer:

  see attached

Explanation:

Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.

Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.

__

Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.

On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.

7 0
2 years ago
What is charge on the cation SnCl4
Rasek [7]
Its total charge is zero but for the elements:
Sn===> Sn4+ positive
Cl===> Cl- negative
7 0
1 year ago
Read 2 more answers
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
1 year ago
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