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

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 sensible heating. Hence, the heat released by the fuel is equal to the heat absorbed by the water because of principle of energy conservation. The heat released 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 grams of P₂O₅ to moles of P₂O₅.

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

Step 2. Convert moles of P₂O₅ to molecules of P₂O₅.

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

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

Step 3. Convert molecules of P₂O₅ to atoms.

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]

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.

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