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2 years ago

What is the density of a substance that has a mass of 90 g and a volume of 18 cm3?

Chemistry

1 answer:

LekaFEV [45]2 years ago

7 0

Answer:

$\boxed{\tt d=5g/cm^3}$

Explanation:

The density of a substance can be found using the following formula.

$d=\frac{m}{v}$

where $m$ is the mass and $v$ is the volume.

The mass of this substance is 90 grams and the volume is 18 cubic centimeters.

$m= 90 g \\v= 18 cm^3$

Substitute the values into the formula.

$d=\frac{90 g}{18 cm^3}$

Divide 90 grams by 18 cubic centimeters.

$d=5 g/cm^3$

The density of the substance is 5 grams per cubic centimeter.

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Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir

nikdorinn [45]

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

An irradiated food is one in which ionizing radiations have been employed to improve food quality.
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If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.

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1 year ago

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =

IgorC [24]

According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2

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2 years ago

Use the conversion factor 1amu=6.66054 x 10^-24 to answer the following questions.. . a) 1.674 x 10^-24 g of neutrons is how man

Wewaii [24]

A.) The conversion factor is 1 amu = $1.66054 ^{-24}g$

To know how many amu in $1.674x10^{-24}g$ grams of neutrons:
$1.674x10^{-24}g ( \frac{x (amu) }{ 1.66054x10^{-24} g)}$

=1.00811 amu

b.) The mass in grams of one lithium ion which has an atomic weight of 6.94 amu.

$6.94 amu (\frac{6.66054 ^{-24} g}{1 amu})$ = $4.62241x10 ^{23} g$

c.) How many amu in 6.492x10^-23g potassium?

$6.492x10^{-23}g ( \frac{x (amu) }{ 1.66054x10^{-24}g })$ = 39.0957 amu

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2 years ago

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Large sharks eat many other marine life animals. The sharks and the animals they eat are all part of which level of organization

Sladkaya [172]

Answer:

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