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Nady [450]
3 years ago
12

A chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume chang

e is 6.9 mL. If the metal weighs 4.9 g, what is the density of the metal?
Chemistry
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

The density of the metal is 0.561 g/mL

Explanation:

The computation of the density of the metal is shown below;

As we know that

The Density of the metal is

= \frac{mass}{volume}

where,

Mass = 4.9g

Change in volume = 6.9 mL

Now place these values to the above formula

So, the density of the metal is

= \frac{4.9g}{6.9mL}

= 0.561 g/mL

Hence, the density of the metal is 0.561 g/mL

We simply applied the above formula so that the correct density could arrive

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Calcium carbide ( CaC2) reacts with water to produce acetylene (C2H2) as shown in the unbalanced reaction below: CaC2(s)+H2O(g)-
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Answer:
46.3g H2O

Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)

then use factor label method to solve

82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
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What is the difference between a homogeneous and a heterogeneous mixture?
Ne4ueva [31]

Answer and explanation :

DIFFERENCE BETWEEN HOMOGENEOUS AND HETEROGENEOUS MIXTURE :

  • In homogeneous mixture the component which are present in the mixture all are in the same proportion whereas in heterogeneous mixture the component which are present in the mixture don't have the same proportion
  • We can not pick out the component of homogeneous mixture but in heterogeneous mixture we can pick out the components from the mixture
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7 0
3 years ago
Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen
Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen=n_1

Moles of acetylene =n_2

\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}

Volume of large tank with oxygen gas, V_1 = 5.00 L

Pressure of the oxygen gas inside the tank = P_1=135 atm

RT=\frac{P_1V_1}{n_1} ..[1]

Volume of small tank with acetylene gas ,V_2= 3.50 L

Pressure of the acetylene gas inside the tank = P_2=?

RT=\frac{P_2V_1}{n_2} ..[2]

Considering both the gases having same temperature T, [1]=[2]

\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}

P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}

=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm

77.14 atm of pressure should be of an acetylene in the tank.

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3 years ago
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The correct answer is b
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