A package of chips weighs 45 grams. How much is that in kilograms?

In baking biscuits and other quick breads, the

stiv31 [10]

Answer:

chemical

Explanation:

The trapped carbon dioxide makes the dough rise, and the alcohol evaporates during the baking process. This is an irreversible chemical change, because by consuming the sugar, the yeast has created new substances

6 0

3 years ago

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A thermometer that was used in a calorimeter was consistently 2.7 °c too high. how would this affect the calculations of δh for

vodomira [7]

The calorimetry experiment is usually done to determine the heat capacity of the sample. The working equation would be:

Q = mCpΔT
Q is the energy
m is the mass of sample
Cp is the heat capacity
ΔT is the temperature

So, if the thermometer is too high, then that would affect ΔT, which would make it greater. Consequently, you would calculate a much lesser heat capacity of the sample compared to the theoretical value.

7 0

3 years ago

Which nonmetal has the greatest potential to be oxidized

Tasya [4]

Answer:

Hydrogen

Explanation:

A reducing agent is a substance which gives up its electrons to become oxidized. Generally, metals are oxidized (reducing agents) while non-metals are reduced (oxidizing agents).

However, hydrogen which is a non-metal is usually oxidized in the presence of stronger oxidizing non-metals such as fluorine and oxygen.

Hydrogen thus, acts as a reducing agent by giving up its electrons to become oxidized. Even though among all non-metals, Hydrogen has the greatest potential to be oxidized, it is a poor reducing agent compared to reactive metals.

7 0

3 years ago

The highly reactive elements in group 7A are known for forming salts. What are they called? transition metals metalloids halogen

liberstina [14]

The group 7A elements are called Halogens.

Please mark as brainliest if this helped! :)

4 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago