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storchak [24]
3 years ago
8

What is a tester? pls answerrr​

Chemistry
1 answer:
cricket20 [7]3 years ago
7 0

a tester is a person who test something especially a new products

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What are 3 ways living creatures use carbohydrates
Zarrin [17]
Food, Storage, and Structure
4 0
3 years ago
In each of the following diatomic molecules, which end of the molecule is positive relative to the other end?
docker41 [41]

Answer:

A- Hydrogen Fluoride, HF

3 0
3 years ago
The following data was collected when a reaction was performed experimentally in the laboratory. Determine the ammount of NaNO3
zepelin [54]

equalize the coefficients:

Al(NO3)3+3NaCl=3NaNO3+AlCl3

according to the equation 1 mol Al(NO3)3 reacted with 3 moles NaCl. hence, 4 moles Al(NO3)3 should react with 12 moles NaCl, and 9 moles reacted NaCl. So NaCl is limiting, we decide further on this reagent.

\frac{9}{3}=\frac{x}{3}

3

9

=

3

x

x=9

9 moles NaNO3

6 0
3 years ago
Magnesium chloride is an important coagulant used in the preparation of tofu from soy milk. Its solubility in water at 200C is 5
Gennadij [26K]

Answer:

(a) Homogeneous. 4.7 g of MgCl₂.

(b) 9.1 g

Explanation:

(a)

At 200°C, we can dissolve 54.6g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:

38.2gWater.\frac{54.6gMgCl_{2}}{100gWater} =20.9g

Since we can dissolve up to 20.9 g of MgCl₂ and we added only 16.2 g, the mixture is homogeneous and we could add 20.9 g -16.2 g = 4.7 g of solute to make it saturated.

(b)

At 800°C, we can dissolve 66.1 g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:

38.2gWater.\frac{66.1gMgCl_{2}}{100gWater} =25.3g

Since we can dissolve up to 25.3 g of MgCl₂ and we added only 16.2 g, we could add 25.3 g - 16.2 g = 9.1 g of solute to make it saturated.

7 0
3 years ago
Calculate ΔHrxn for the following reaction: Fe₂O₃(s)+3CO(g)→2Fe(s)+3CO₂(g) Use the following reactions and given ΔH′s. 2Fe(s)+3/
ExtremeBDS [4]

Answer : The enthalpy change of reaction is -23.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2Fe(s)+\frac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)    \Delta H^o_1=-824.2kJ

(2) CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)    \Delta H^o_2=-282.7kJ

First we will reverse the reaction 1 and multiply equation 2 by 3 then adding both the equation, we get :

(1) Fe_2O_3(s)\rightarrow 2Fe(s)+\frac{3}{2}O_2(g)    \Delta H^o_1=+824.2kJ

(2) 3CO(g)+\frac{3}{2}O_2(g)\rightarrow 3CO_2(g)    \Delta H^o_2=3\times (-282.7kJ)=-848.1kJ

The expression for final enthalpy is,

\Delta H=\Delta H^o_1+\Delta H^o_2

\Delta H=(+824.2kJ)+(-848.1kJ)

\Delta H=-23.9kJ

Therefore, the enthalpy change of reaction is -23.9 kJ

7 0
3 years ago
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