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beks73 [17]
3 years ago
13

10.0 g of Mg(NO3)2 = ____ moles

Chemistry
1 answer:
JulsSmile [24]3 years ago
6 0

Answer:

0.067424154568526 moles

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Which statement is true of chloroplasts?
Anastasy [175]

d answer is correct that help

5 0
2 years ago
The heat of combustion of ethane is -337.0kcal at 25 degrees celsius. what is the heat of the reaction when 3g of ethane is burn
Ghella [55]

Answer:

Q = -33.6kcal .

Explanation:

Hello there!

In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

Q=n*\Delta _cH

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

n=3g*\frac{1mol}{30.08g}=0.1mol

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

Q=0.1mol*-337.0\frac{kcal}{mol}\\\\Q=-33.6kcal

Best regards!

8 0
2 years ago
The electron affinity of Bi is -94.6 kJ and the value for At is -280 kJ. What is the approximate electron affinity for Po?
lys-0071 [83]

Answer:

48 KJ mol-1

Explanation:

3 0
2 years ago
What is the bond angle in a tetrahedral molecule?
levacccp [35]
109.5

tetrahedral shape:
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7 0
3 years ago
Read 2 more answers
1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced
koban [17]

Answer:

\large \boxed{\text{200 g CO}_{{2}}}

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ:                                 44.01

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:    1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol  CO₂:1 mol C₃H₈

\rm  \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}

2. Calculate the mass of CO₂.

\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2}  \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}

3 0
2 years ago
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