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Tanzania [10]
2 years ago
14

Density = 7 g/mL Volume = 10mL ∙What is the Mass?

Chemistry
1 answer:
Mrrafil [7]2 years ago
7 0

Answer:

Mass is 70

Explanation:

Density= Mass divided by Volume

To find the mass you do the opposite of the equation- Volume times the density. Therefore, the mass is 70 because 10 times 7 is 70.

Hope this helps :)

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If you place rice on top or a speaker, what will happen to the rice when you turn the speaker on?
Dovator [93]

Answer:

The rice will move

Explanation:

the sound vibration and waves will cause the rice to move

3 0
2 years ago
How many grams of Cl are in 465g of CaCl2
Rama09 [41]
2 ways to do this
a. find %Cl in CaCl2
2 x 35.45g/mole = 70.9g Cl
70.9g Cl / 110.9g/mole CaCl2 = 63.93% Cl in CaCl2
0.6963 x 145g = 92.7g = mass Cl

b. determine moles CaCl2 present then mass Cl
145g / 110.9g/mole = 1.31moles CaCl2 present
2moles Cl / 1mole CaCl2 x 1.31moles = 2.62moles Cl
2.62moles Cl x 35.45g/mole = 92.7g Cl
6 0
3 years ago
Read 2 more answers
In an exothermic reaction, the bonding energy of the product is:
ahrayia [7]
I think the correct answer from the choices listed above is option A. <span>In an exothermic reaction, the bonding energy of the product is </span><span>less than the reactant because it is only at this condition that the energy is released by the reaction.</span>
3 0
3 years ago
Read 2 more answers
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
2 years ago
3. Indicate the type of reaction represented by each equation:
BartSMP [9]
3. Is double replacement
6 0
3 years ago
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