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3 years ago

Which substance is a mixture Table salt Gasoline Aluminum Carbon Dioxide

Chemistry

2 answers:

NNADVOKAT [17]3 years ago

8 0

Answer:

carbon dioxide

Explanation:

its 2 or more things

evablogger [386]3 years ago

7 0

Answer: Gasoline
Explanation: a mixture is a mixture of two separate compounds. Table salt is Sodium Chloride which is only one compound (NaCl), carbon dioxide is only one compound (CO2), and aluminum is an element on the periodic table so it isn’t a mixture of anything, Gasoline on the other hand is a mixture of multiple different compounds.

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Molarity of 0.50 mol sugar in 270 mL of solution.

EastWind [94]

<h3>Answer:</h3>

1.85 M

<h3>Explanation:</h3>

<u>We are given;</u>

Number of moles as 0.50 mol
Volume of the solution is 270 ml

But, 1000 mL = 1 L

Thus, volume of the solution is 0.27 L

We are required to calculate the molarity of the solution;

Molarity refers to the concentration of a solution in moles per liter.
It is calculated by dividing number of moles with the volume.

Molarity = Moles ÷ Volume

In this case;

Molarity = 0.50 moles ÷ 0.27 L

= 1.85 Mol/L or 1.85 M

Therefore, molarity of the solution is 1.85 M

5 0

2 years ago

1.Calculate the molarity of a 6.15% (w/v%) sodium hypochlorite (NaOCl) solution (i.e. in a 100 ml of solution there is 6.15 gram

shtirl [24]

Answer:

C U M

Explanation:

3 0

2 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

2 years ago

What is the percent composition by mass of sulfur in the compound mgso4

Margaret [11]

Answer:

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

Explanation:

Molar mass of compound = 120 g/mol

Number of sulfur atom = 1

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

$=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100$

Sulfur :

$=\frac{1\times 32 g/mol}{120 g/mol}\times 100=26.7\%$

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

6 0

3 years ago

Which has not been a major source of CFCs

ipn [44]

Answer : Any natural sources of CFC's are not known only the major sources like aerosols, propellants, refrigerants,etc are known. So, if any natural sources are given then it cannot be called as a major source for emitting CFC into environment.

4 0

3 years ago

Read 2 more answers