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xenn [34]
3 years ago
14

How to make baking powder

Chemistry
2 answers:
Keith_Richards [23]3 years ago
6 0

Answer:

To make baking powder, combine half a teaspoon of cream of tartar and a quarter teaspoon of bicarbonate of soda.

hope it helps;)

deff fn [24]3 years ago
4 0

Answer:

with baking soda

Explanation:

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Among these element groups, which is least likely to form ions?
cupoosta [38]

Your answer would be 7A

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3 years ago
How much heat is needed to melt 100.0 grams of ice that is already at 0°C?
Rashid [163]

A. The heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.

<h3>What is Specific heat capacity?</h3>

Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass.

<h3>Heat needed to melt the cube of ice</h3>

The heat is needed to melt 100.0 grams of ice that is already at 0°C is calculated as follows;

Q = mL

where;

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Q = 100 x 334

Q = 33,400 J

Thus, the heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.

Learn more about heat capacity here: brainly.com/question/16559442

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6 0
1 year ago
Which planet has a distinctive red spot? A. Mercury B. Venus C. Jupiter D. Mars
Firdavs [7]
Jupiter------------------------
3 0
3 years ago
Read 2 more answers
A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus
SIZIF [17.4K]
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal

<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>
8 0
3 years ago
What is the mass of 1.7 × 1023 atoms of zinc
fiasKO [112]

We are given with a compound, Zinc (Zn) having a 1.7 x 10 ^23 atoms. We are tasked to solve for it's corresponding mass in g. We need to find first the molecular weight of Zinc, that is

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Not that 1 mol=6.022x10^{23} atoms, hence,

1.7 x 10 ^23 atoms x 1 mol/6.022x10^{23} atoms x65.38 g/ 1mol

=18.456 g of Zn

 

Therefore, the mass of Zinc 18.456 g

3 0
3 years ago
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