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Kay [80]
3 years ago
12

Is diphenylamine a solid, liquid, or a gas at room temperature

Chemistry
1 answer:
goblinko [34]3 years ago
3 0

Answer:

Solid

Explanation:

Diphenylamine has a melting point of 127.4 F or 53 C so at room temperature ~70 F or 21 C its a solid.

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A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea
jekas [21]

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:

Q = m\cdot c \cdot \Delta T (1)

Where:

  • m - Mass of the sample, in grams.
  • c - Specific heat of water, in calories per gram-degree Celsius.
  • \Delta T - Temperature change, in degrees Celsius.

If we know that m = 10\,g, c = 1\,\frac{cal}{g\cdot ^{\circ}C} and \Delta T = 9\,^{\circ}C, then the heat released by the fuel is:

Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

7 0
2 years ago
7. How many liters of NH3, at STP will react with 10.6 g O2 to form NO2 and water? 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) *
Alexus [3.1K]

7) Answer is: c. 4.24 L.

Balanced chemical reaction: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).

m(O₂) = 10.6 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 10.6 g ÷ 32 g/mol.

n(O₂) = 0.33 mol; amount of oxygen.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

V(O₂) = n(O₂) · Vm.

V(O₂) = 0.33 mol · 22.4 L/mol.

V(O₂) = 7.42 L; volume of oxygen.

From balanced chemical reaction: n(O₂) : n(NH₃) = 7 : 4.

n(NH₃) = 4 · 0.33 mol ÷ 7.

n(NH₃) = 0.188 mol; amount of ammonia.

V(NH₃) = 0.188 mol · 22.4 L/mol.

V(NH₃) = 4.24 L.

8) Answer is: a. 3.7 g.

Balanced chemical reaction: P₄(g) + 6H₂(g) → 4PH₃(g).

m(P₄) = 3.4 g; mass of phosphorous.

n(P₄) = m(P₄) ÷ M(P₄).

n(P₄) = 3.4 g ÷ 123.9 g/mol.

n(P₄) = 0.0274 mol; limiting reactant.

n(H₂) = 4 g ÷ 2 g/mol.

n(H₂) = 2 mol; amount of hydrogen.

From balanced chemical reaction: n(P₄) : n(PH₃) = 1 : 4.

n(PH₃) = 4 · 0.0274 mol.

n(PH₃) = 0.1096 mol.

m(PH₃) = 0.1096 mol · 34 g/mol.

m(PH₃) = 3.726 g.

9) Answer is: d. Percent yield is the ratio of theoretical yield to actual yield expressed as a percent.

Percent yield = actual yield / theoretical yield.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

For example:

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.

3 0
3 years ago
How many total atoms are in 0.290 g of P2O5?
Sliva [168]

There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\

= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\

Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

∴ \text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\

8 0
3 years ago
Which resources are significantly polluted by heavy metals like mercury?
arlik [135]
<span>Heavy metals like mercury enter waterways by industrial dumping and poor regulatioin of effluent, and they also enter soil through a similar manner, in which waste is disposed of imporperly. Another source of heavy metals are the gases leaving industry carrying these metals. The metals fall as a solid on to soil and water ways. Therefore, the answer is D.</span>
4 0
3 years ago
Read 2 more answers
Which compound has the same empirical and molecular formula ethyne ethene ethane methane?
Margaret [11]
Empirical formula is the simplest ratio of whole numbers of components in a compound 
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne  - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements                 simplified ratio
                     C : H                                           C : H
ethyne            2:2                                             1:1
ethene            2:4                                             1:2
ethane            2:6                                             1:3
methane         1:4                                             1:4
the only compound where the actual ratio is equal to the simplified ratio is methane 
therefore in methane molecular formula CH₄ is the same as empirical formula CH₄
6 0
3 years ago
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